\Big{(}\Gamma(z/\beta)\Big{)}^{\mu_{1}}\prod\limits_{n=1}^{+\infty}\frac{\Gamma\Big{(}\frac{z-\mathfrak{z}_{-n}}{\beta}\Big{)}}{\Gamma\Big{(}\frac{\mathfrak{z}_{n}-z}{\beta}+1\Big{)}}\Big{[}\frac{\mathfrak{z}_{n}(-\mathfrak{z}_{-n})}{\beta^{2}}\Big{]}^{\frac{\beta-2z}{2\beta}}\exp\\{\mathcal{R}^{I}\\},\quad\text{if
}\mathcal{S}_{1}\text{ has an infinite number }\\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{of
zeros and }\beta>0,\\\
\Big{(}\Gamma(z/\beta)\Big{)}^{\mu_{1}}\prod\limits_{n=1}^{+\infty}\frac{\Gamma\Big{(}\frac{\mathfrak{z}_{n}-z}{|\beta|}\Big{)}}{\Gamma\Big{(}\frac{z-\mathfrak{z}_{-n}}{|\beta|}+1\Big{)}}\Big{[}\frac{\mathfrak{z}_{n}(-\mathfrak{z}_{-n})}{\beta^{2}}\Big{]}^{\frac{\beta-2z}{2\beta}}\exp\\{\mathcal{R}^{I}\\},\quad\text{if
}\mathcal{S}_{1}\text{ has an infinite number}\\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad\text{
of zeros and }\beta<0,\end{cases}$ $\displaystyle\ \mathbb{L}^{II}(z)$
$\displaystyle=\begin{cases}\Big{(}\Gamma\big{(}\frac{z}{\beta}\big{)}\Big{)}^{-\mu_{2}}\frac{\prod\limits_{n=1}^{N_{3}}\Gamma\Big{(}\frac{\mathfrak{y}_{n}-z}{\beta}+1\Big{)}}{\prod\limits_{n=1}^{N_{4}}\Gamma\Big{(}\frac{z-\mathfrak{y}_{-n}}{\beta}\Big{)}}\exp\\{i\pi
z(N_{3}+1)\\},\quad\text{if }\mathcal{S}_{2}\text{ has a finite number of
zeros},\\\
\Big{(}\Gamma(z/\beta)\Big{)}^{-\mu_{2}}\prod\limits_{n=1}^{+\infty}\frac{\Gamma\Big{(}\frac{\mathfrak{y}_{n}-z}{\beta}+1\Big{)}}{\Gamma\Big{(}\frac{z-\mathfrak{y}_{-n}}{\beta}\Big{)}}\Big{[}\frac{\mathfrak{y}_{n}(-\mathfrak{y}_{-n})}{\beta^{2}}\Big{]}^{\frac{2z-\beta}{2\beta}}\exp\\{\mathcal{R}^{II}\\},\quad\text{if
}\mathcal{S}_{2}\text{ has an infinite number }\\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{
of zeros and }\beta>0,\\\
\Big{(}\Gamma(z/\beta)\Big{)}^{-\mu_{2}}\prod\limits_{n=1}^{+\infty}\frac{\Gamma\Big{(}1+\frac{z-\mathfrak{y}_{-n}}{|\beta|}\Big{)}}{\Gamma\Big{(}\frac{\mathfrak{y}_{n}-z}{|\beta|}\Big{)}}\Big{[}\frac{\mathfrak{y}_{n}(-\mathfrak{y}_{-n})}{\beta^{2}}\Big{]}^{\frac{2z-\beta}{2\beta}}\exp\\{\mathcal{R}^{II}\\},\quad\text{if
}\mathcal{S}_{1}\text{ has an infinite number }\\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{
of zeros and }\beta<0.\end{cases}$

Finally, denoting

$\mathfrak{d}_{0}^{\star}=\begin{cases}A_{0}^{I}\beta^{-\mu_{1}}\quad\text{in
the \bf{FFK} case},\\\ \,\\\ \frac{\beta^{\mu_{2}}}{A_{0}^{II}}\quad\text{in
the \bf{FSK} case},\\\ \,\\\
\frac{\beta^{\mu_{2}-\mu_{1}}A_{0}^{I}}{A_{0}^{II}}\quad\text{in the \bf{FTK}
case},\end{cases}\quad B^{\star}=\begin{cases}B_{1}^{I}\quad\quad\text{in the
\bf{FFK} case},\\\ \,\\\ B_{1}^{II}\quad\quad\text{in the \bf{FSK} case},\\\
\,\\\ B_{1}^{I}-B_{2}^{II}\quad\text{in the \bf{FTK} case},\end{cases}$
$\mathbb{L}^{\star}(z)=\begin{cases}\mathbb{L}^{I}\quad\quad\text{in the
\bf{FFK} case},\\\ \,\\\ \mathbb{L}^{II}\quad\quad\text{in the \bf{FSK}
case},\\\ \,\\\ \frac{\mathbb{L}^{I}}{\mathbb{L}^{II}}\quad\text{in the
\bf{FTK} case},\end{cases}$

and appealing to Theorem 2.1, we end up with the assertion.

######  Theorem 5.1.

Let assumptions H1 and H8-H10 hold, $\mathbb{F}\equiv 0$. Then a general
solution $Y_{h}(z,\sigma;{\mathbb{P}})$ of homogenous equation (5.1) is given
by

$Y_{h}(z,\sigma;{\mathbb{P}})=\exp\Big{\\{}\frac{B^{\star}}{2\beta}z^{2}-\frac{B^{\star}}{2}z\Big{\\}}(\mathfrak{d}_{0}^{\star})^{\frac{2z-\beta}{2\beta}}{\mathbb{P}}(z/\beta)(a_{1}\sigma+a_{2}\sigma)^{\frac{\beta-2z}{2\beta}}\mathbb{L}^{\star}(z),$
(5.13)

where ${\mathbb{P}}(z)$ is a function describing in Theorem 2.1.

Moreover, statements (s.1), (s.2) of Theorem 2.1 with
$\mathbb{L}(z):=\mathbb{L}^{\star}(z)$ and
$\mathbb{L}_{1}(z):=\mathbb{L}^{\star}(z)$ hold.

Coming to inhomogeneous equation (5.1), we state additional hypotheses.

H12: For $d_{0}\in[0,1]$ and $m\in\mathbb{N}\cup\\{0\\},$ we assume

$Re\Big{(}\frac{z-\mathfrak{z}_{n}}{\beta}\Big{)}\neq d_{0}+m,\quad
n\in\\{1,2,...,N_{1}\\},$

in the case of a finite number of zeros at $\mathcal{S}_{1}$. Otherwise, we
require that these inequalities hold with $n\in\mathbb{N}$ if $\beta>0$, while
in the case of $\beta<0$ the inequalities are fulfilled

$Re\Big{(}\frac{z-\mathfrak{z}_{-n}}{|\beta|}\Big{)}\neq-d_{0}-m,\quad
n\in\mathbb{N}.$

In the case of a finite number of zeros at $\mathcal{S}_{2}$, we require that

$Re(z/\beta)\neq-1-m+d_{0}\quad\text{if}\quad\mu_{2}\neq 0;\quad\text{and
}Re\Big{(}\frac{z-\mathfrak{y}_{-n}}{\beta}\Big{)}\neq-m-1+d_{0},\quad
n\in\\{1,2,...,N_{4}\\}.$

Otherwise, we require that these relations hold with $n\in\mathbb{N}$ if
$\beta>0$, while in the case of $\beta<0$ the inequalities are fulfilled

$Re(z/\beta)\neq-1-m+d_{0}\quad\text{if}\quad\mu_{2}\neq 0;\quad\text{and
}Re\Big{(}\frac{z-\mathfrak{y}_{n}}{|\beta|}\Big{)}\neq m+1-d_{0},\quad
n\in\mathbb{N}.$

In the case of the function $\mathcal{S}_{1}/\mathcal{S}_{2}$, we assume that
all inequalities hold simultaneously.

To make the following assumption, we set
${\mathbb{P}}_{1}:={\mathbb{P}}(z/\beta)$ for each chosen periodic function
and $Y_{h}(z,\sigma)=Y_{h}(z,\sigma;{\mathbb{P}}_{1}).$

H13 We assume that $\mathbb{F}(z,\sigma)$ is an analytic function for
$z\in{\mathbb{C}}$ and each $\sigma$ satisfying H1. Moreover, for some
functions ${\mathbb{P}}_{1}(z/\beta)$ and $\mathcal{K}_{1}(\xi)$, the equality
holds

$\underset{|Im\xi|\to+\infty}{\lim}\Big{|}\frac{\mathcal{K}(\xi)\mathbb{F}(z+\beta\xi,\sigma)}{Y_{h}(z+\beta\xi+\beta,\sigma)}\Big{|}=0.$
(5.14)

for all $z$ satisfying H.11 and H.12 and each $Re\xi\in[-1,0]$.

Now, taking into account Theorems 2.2, 4.1 and 5.1 we conclude.

######  Theorem 5.2.

Let assumptions of Theorem 5.1 and hypotheses H5, H12, H13 hold. Then a
general solution of (5.1) has a form
$Y_{g}(z,\sigma)=Y_{h}(z,\sigma;{\mathbb{P}})+Y_{ih}(z,\sigma),$ where
$Y_{h}(z,\sigma;{\mathbb{P}})$ is defined in Theorem 5.1 and
$Y_{ih}(z,\sigma)$ being a particular solution is given by

$Y_{ih}(z,\sigma)=\frac{Y_{h}(z,\sigma)}{\mathcal{K}_{1}(0)}\int\limits_{\ell_{d_{0}}}\frac{\mathbb{F}(z+\beta\xi,\sigma)\mathcal{K}(\xi)}{(a_{1}\sigma+a_{2}\sigma^{\nu})Y_{h}(z+\beta\xi+\beta,\sigma)}d\xi.$
(5.15)

Finally, to demonstrate results of Theorems 5.1-5.2, we conclude this section
with examples of explicit solutions to (5.1) with the coefficient
$\mathcal{S}$ presented in Propositions 5.1-5.3. In light of formulas (5.13)
and (5.15), we give below the explicit forms to $\mathbb{L}^{\star}(z),$
$B^{\star}$, $\mathfrak{d}_{0}^{\star}$ and the kernel $\mathcal{K}(\xi)$. As
for $\mathcal{K}(\xi)$, we describe several kinds of this function which
correspond to the various types of the functions ${\mathbb{P}}(z/\beta)$ and
$\mathbb{F}$. For simplicity consideration, we analyze (5.1) in the case of
$\beta=1$.

We begin with consideration of the homogenous case of (5.1). By virtue of
Propositions 5.1-5.3, we deduce that $B^{\star}\equiv 0$, which means the term
$\exp\\{\frac{B^{\star}z^{2}}{2\beta}-\frac{B^{\star}z}{2}\\}$ equals to $1$
in (5.13). To this end, for $q_{1}^{\star}=\frac{p^{\star}}{2q^{\star}}>1,$
$q_{2}^{\star}>0$, with $p^{\star}$ and $q^{\star}$ having the properties of
the numbers $p,q$, we first denote the zeros of the function
$\mathcal{S}^{\pm}(z;\theta_{1},\theta_{2},q_{1}^{\star},q_{2}^{\star})$ in
the segment $[0,4\pi q^{\star})$ by

$\mathfrak{z}_{i}^{\star,+}:=\mathfrak{z}_{i}^{\star,+}(\theta_{1},\theta_{2},q_{1}^{\star},q_{2}^{\star}),\quad
i\in\\{1,2,...,{\boldsymbol{K}}^{\star,+}\\},\quad\text{and}\quad\mathfrak{z}_{i}^{\star,-}:=\mathfrak{z}_{i}^{\star,-}(\theta_{1},\theta_{2},q_{1}^{\star},q_{2}^{\star}),\quad
i\in\\{1,2,...,{\boldsymbol{K}}^{\star,-}\\},$

where ${\boldsymbol{K}}^{\star,+}$ and ${\boldsymbol{K}}^{\star,-}$ are the
corresponding number of the zeros in $[0,4\pi q^{\star})$. Then, collecting
Theorem 5.1 with Propositions 5.1-5.3, we end up with the outputs listed in
Tables LABEL:tab:table3-LABEL:tab:table2.

Table 2. The quantity of $\mathfrak{d}_{0}^{\star}$ for $\mathcal{S}(z)$ listed in Table LABEL:tab:table2 $\mathcal{S}(z)$ | $\mathfrak{d}_{0}^{\star}$
---|---
$\sin(q_{0}z+\theta),\quad\theta\in[0,2\pi)$ | $(-1)^{\lfloor\theta/\pi\rfloor}q_{0}\frac{\sin\theta_{+}}{\theta_{+}}$
$\cos(q_{0}z-\theta),\,\theta\in[0,2\pi),\,\theta\neq\frac{\pi}{2},\frac{3\pi}{2}$ | $(-1)^{\lfloor\frac{2\theta+\pi}{2\pi}\rfloor}\cos\theta_{-}(\theta)$
$\mathcal{S}^{+}(z;0,0,q_{1},q_{2})$ | $(1+q_{2}q_{1})\prod\limits_{i=1}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i}^{+})^{-2}$
$\mathcal{S}^{-}(z;0,0,q_{1},q_{2}),\,q_{2}\neq\frac{1}{q_{1}}$ | $(1-q_{2}q_{1})\prod\limits_{i=1}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i}^{-})^{-2}$
$\mathcal{S}^{-}(z;0,0,q_{1},q_{2}),\,q_{2}=\frac{1}{q_{1}}$ | $-(1-q_{2}q_{1}^{3})\prod\limits_{i=4}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i}^{-})^{-2}$
$\begin{array}[]{l}\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2}),\,\text{if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}-q_{2}\sin\theta_{2}\neq 0\end{array}$ | $-(\sin\theta_{1}-q_{2}\sin\theta_{2})\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i}^{-})^{-1}$
$\begin{array}[]{l}\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2}),\,\text{if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}-q_{2}\sin\theta_{2}=0\end{array}$ | $(-1)^{\lfloor\mu_{1}^{-}/2\rfloor}[\cos\theta_{1}-q_{2}q_{1}^{\mu_{1}^{-}}\cos\theta_{2}]\prod\limits_{i=\mu_{1}^{-1}+1}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i}^{-})^{-1}$
$\begin{array}[]{l}\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2}),\ \text{ if }\theta_{1},\,\theta_{2}\in(0,\pi)\\\ \sin\theta_{1}+q_{2}\sin\theta_{2}\neq 0\end{array}$ | $-[\sin\theta_{1}+q_{2}\sin\theta_{2}]\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i}^{+})^{-1}$
$\begin{array}[]{l}\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2}),\,\text{if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}+q_{2}\sin\theta_{2}=0\end{array}$ | $(-1)^{\lfloor\mu_{1}^{+}/2\rfloor}[\cos\theta_{1}+q_{2}q_{1}^{\mu_{1}^{+}}\cos\theta_{2}]\prod\limits_{i=\mu_{1}^{+}+1}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i}^{+})^{-1}$
$\begin{array}[]{l}\frac{\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}-q_{2}\sin\theta_{2}=0\end{array}$ | $\frac{(-1)^{\lfloor\mu_{1}^{-}/2\rfloor}(\cos\theta_{1}-q_{2}q_{1}^{\mu_{1}^{-}}\cos\theta_{2})}{(1+q_{2}^{\star}q_{1}^{\star})}\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}(\mathfrak{z}_{i}^{\star,+})^{2}\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i}^{-})^{-1}$
$\begin{array}[]{l}\frac{\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{-}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\,\text{if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}-q_{2}\sin\theta_{2}=0,\,q^{\star}_{2}\neq 1/q_{1}^{\star}\end{array}$ | $\frac{(-1)^{\lfloor\mu_{1}^{-}/2\rfloor}(\cos\theta_{1}-q_{2}q_{1}^{\mu_{1}^{-}}\cos\theta_{2})}{(1-q_{2}^{\star}q_{1}^{\star})}\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,-}-1}(\mathfrak{z}_{i}^{\star,-})^{2}\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i}^{-})^{-1}$
$\begin{array}[]{l}\frac{\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{-}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\,\text{if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}-q_{2}\sin\theta_{2}=0,\,q^{\star}_{2}=1/q_{1}^{\star}\end{array}$ | $\frac{(-1)^{\lfloor\mu_{1}^{-}/2\rfloor-1}(\cos\theta_{1}-q_{2}q_{1}^{\mu_{1}^{-}}\cos\theta_{2})}{(1-q_{2}^{\star}(q_{1}^{\star})^{3})}\prod\limits_{i=4}^{{\boldsymbol{K}}^{\star,-}-1}(\mathfrak{z}_{i}^{\star,-})^{2}\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i}^{-})^{-1}$
$\begin{array}[]{l}\frac{\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}+q_{2}\sin\theta_{2}\neq 0\end{array}$ | $-\frac{(\sin\theta_{1}+q_{2}\sin\theta_{2})}{(1+q_{2}^{\star}q_{1}^{\star})}\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}(\mathfrak{z}_{i}^{\star,+})^{2}\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i}^{+})^{-1}$
$\begin{array}[]{l}\frac{\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}+q_{2}\sin\theta_{2}=0\end{array}$ | $\frac{(-1)^{\lfloor\mu_{1}^{+}/2\rfloor}(\cos\theta_{1}+q_{2}q_{1}^{\mu_{1}^{+}}\cos\theta_{2})}{(1+q_{2}^{\star}q_{1}^{\star})}\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}(\mathfrak{z}_{i}^{\star,+})^{2}\prod\limits_{i=\mu_{1}^{+}+1}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i}^{+})^{-1}$
Table 3. The function $\mathbb{L}^{\star}(z)$ for different kinds of $\mathcal{S}(z)$ $\mathcal{S}(z)$ | $\mathbb{L}^{\star}(z)$
---|---
$\mathcal{S}^{+}(z;0,0,q_{1},q_{2})$ | $\begin{array}[]{l}\Gamma(z)\prod\limits_{i=1}^{{\boldsymbol{K}}^{+}-1}\frac{\Gamma(\mathfrak{z}_{i}^{+}+z)}{\Gamma(\mathfrak{z}_{i}^{+}-z+1)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\frac{\Gamma(\mathfrak{z}_{i,n}^{+}+z)}{\Gamma(\mathfrak{z}_{i,n}^{+}-z+1)}[\mathfrak{z}_{i,n}^{+}]^{1-2z}\end{array}$
$\mathcal{S}^{-}(z;0,0,q_{1},q_{2}),\,q_{2}\neq\frac{1}{q_{1}}$ | $\begin{array}[]{l}\Gamma(z)\prod\limits_{i=1}^{{\boldsymbol{K}}^{-}-1}\frac{\Gamma(\mathfrak{z}_{i}^{-}+z)}{\Gamma(\mathfrak{z}_{i}^{-}-z+1)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\frac{\Gamma(\mathfrak{z}_{i,n}^{-}+z)}{\Gamma(\mathfrak{z}_{i,n}^{-}-z+1)}[\mathfrak{z}_{i,n}^{-}]^{1-2z}\end{array}$
$\mathcal{S}^{-}(z;0,0,q_{1},q_{2}),\,q_{2}=\frac{1}{q_{1}}$ | $\begin{array}[]{l}[\Gamma(z)]^{3}\prod\limits_{i=4}^{{\boldsymbol{K}}^{-}-1}\frac{\Gamma(\mathfrak{z}_{i}^{-}+z)}{\Gamma(\mathfrak{z}_{i}^{-}-z+1)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\frac{\Gamma(\mathfrak{z}_{i,n}^{-}+z)}{\Gamma(\mathfrak{z}_{i,n}^{-}-z+1)}[\mathfrak{z}_{i,n}^{-}]^{1-2z}\end{array}$
$\begin{array}[]{l}\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2}),\text{if }\theta_{1},\,\theta_{2}\in(0,\pi)\\\ \text{and }\sin\theta_{1}-q_{2}\sin\theta_{2}\neq 0\end{array}$ | $\begin{array}[]{l}\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\frac{1}{\Gamma(\mathfrak{z}_{i}^{-}-z+1)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\frac{\Gamma(z-\mathfrak{z}_{i,-n}^{-})[(-\mathfrak{z}_{i,-n}^{-})\mathfrak{z}_{i,n}^{-}]^{\frac{1-2z}{2}}}{\Gamma(\mathfrak{z}_{i,n}^{-}-z+1)}\\\ \times\exp\\{\mathfrak{z}_{i,-n}^{-}[\ln(-\mathfrak{z}_{i,-n}^{-})-1]+\mathfrak{z}_{i,n}^{-}[\ln(\mathfrak{z}_{i,n}^{-})-1]\\}\end{array}$
$\begin{array}[]{l}\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2}),\text{if }\theta_{1},\,\theta_{2}\in(0,\pi)\\\ \text{and }\sin\theta_{1}-q_{2}\sin\theta_{2}=0\end{array}$ | $\begin{array}[]{l}\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}\frac{[\Gamma(z)]^{\mu_{1}^{-}}}{\Gamma(\mathfrak{z}_{i}^{-}-z+1)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\frac{\Gamma(z-\mathfrak{z}_{i,-n}^{-})[(-\mathfrak{z}_{i,-n}^{-})\mathfrak{z}_{i,n}^{-}]^{\frac{1-2z}{2}}}{\Gamma(\mathfrak{z}_{i,n}^{-}-z+1)}\\\ \times\exp\\{\mathfrak{z}_{i,-n}^{-}[\ln(-\mathfrak{z}_{i,-n}^{-})-1]+\mathfrak{z}_{i,n}^{-}[\ln(\mathfrak{z}_{i,n}^{-})-1]\\}\end{array}$
$\begin{array}[]{l}\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2}),\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi)\\\ \sin\theta_{1}+q_{2}\sin\theta_{2}\neq 0\end{array}$ | $\begin{array}[]{l}\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\frac{1}{\Gamma(\mathfrak{z}_{i}^{+}-z+1)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\frac{\Gamma(z-\mathfrak{z}_{i,-n}^{+})[(-\mathfrak{z}_{i,-n}^{+})\mathfrak{z}_{i,n}^{+}]^{\frac{1-2z}{2}}}{\Gamma(\mathfrak{z}_{i,n}^{+}-z+1)}\\\ \times\exp\\{\mathfrak{z}_{i,-n}^{+}[\ln(-\mathfrak{z}_{i,-n}^{+})-1]+\mathfrak{z}_{i,n}^{+}[\ln(\mathfrak{z}_{i,n}^{+})-1]\\}\end{array}$
$\begin{array}[]{l}\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2})\,\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi)\\\ \text{and}\,\sin\theta_{1}+q_{2}\sin\theta_{2}=0\end{array}$ | $\begin{array}[]{l}\prod\limits_{i=\mu_{1}^{+}+1}^{{\boldsymbol{K}}^{+}-1}\frac{[\Gamma(z)]^{\mu_{1}^{+}}}{\Gamma(\mathfrak{z}_{i}^{+}-z+1)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\frac{\Gamma(z-\mathfrak{z}_{i,-n}^{+})[(-\mathfrak{z}_{i,-n}^{+})\mathfrak{z}_{i,n}^{+}]^{\frac{1-2z}{2}}}{\Gamma(\mathfrak{z}_{i,n}^{+}-z+1)}\\\ \times\exp\\{\mathfrak{z}_{i,-n}^{+}[\ln(-\mathfrak{z}_{i,-n}^{+})-1]+\mathfrak{z}_{i,n}^{+}[\ln(\mathfrak{z}_{i,n}^{+})-1]\\}\end{array}$
$\sin(q_{0}z+\theta),$ if $\theta\in[0,2\pi)$ | $\begin{array}[]{l}\Gamma\Big{(}z+\frac{\theta_{+}}{q_{0}}\Big{)}\prod\limits_{n=1}^{+\infty}\frac{\Gamma\Big{(}\frac{\pi n-\theta_{+}}{q_{0}}+z\Big{)}}{\Gamma\Big{(}\frac{\pi n+\theta_{+}}{q_{0}}-z+1\Big{)}}\Big{(}\frac{\pi^{2}n^{2}-\theta_{+}^{2}}{q_{0}^{2}}\Big{)}^{\frac{1}{2}-z}\\\ \times\exp\Big{\\{}\frac{\pi n-\theta_{+}}{q_{0}}\Big{[}1-\ln\frac{\pi n-\theta_{+}}{q_{0}}\Big{]}+\frac{\pi n+\theta_{+}}{q_{0}}\Big{[}\ln\frac{\pi n+\theta_{+}}{q_{0}}-1\Big{]}\Big{\\}}\end{array}$
$\cos(q_{0}z-\theta)$, if $\theta\in[0,2\pi),\,\theta\neq\frac{\pi}{2},\frac{3\pi}{2}$ | $\begin{array}[]{l}\prod\limits_{n=1}^{+\infty}\frac{\Gamma\Big{(}\frac{\pi(2n-1)-2\theta_{-}}{2q_{0}}+z\Big{)}}{\Gamma\Big{(}\frac{\pi(2n-1)+2\theta_{-}}{2q_{0}}-z+1\Big{)}}\Big{(}\frac{\pi^{2}(2n-1)^{2}-4\theta_{-}^{2}}{4q_{0}^{2}}\Big{)}^{\frac{1}{2}-z}\\\ \times\exp\Big{\\{}\frac{\pi(2n-1)-2\theta_{-}}{2q_{0}}\Big{[}1-\ln\frac{\pi(2n-1)-2\theta_{-}}{2q_{0}}\Big{]}\\\ +\frac{\pi(2n-1)+2\theta_{-}}{2q_{0}}\Big{[}\ln\frac{\pi(2n-1)+2\theta_{-}}{2q_{0}}-1\Big{]}\Big{\\}}\end{array}$
$\begin{array}[]{l}\frac{\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\,\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi)\\\ \text{and }\sin\theta_{1}-q_{2}\sin\theta_{2}=0\end{array}$ | $\begin{array}[]{l}\frac{[\Gamma(z)]^{\mu_{1}^{-}-1}\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}\Gamma(\mathfrak{z}_{i}^{-}-z+1)}{\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i}^{\star,+}+z)\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}\Gamma(\mathfrak{z}_{i}^{-}-z+1)}\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{-})}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{-}+1)}\\\ \times\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,+}-z+1)}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,+}+z)}\Big{[}\frac{(-\mathfrak{z}_{i,-n}^{-})\mathfrak{z}_{i,n}^{-}}{(\mathfrak{z}_{i,n}^{\star,+})^{2}}\Big{]}^{\frac{1-2z}{2}}\\\ \times\exp\Big{\\{}\sum\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i,-n}^{-}[\ln(-\mathfrak{z}_{i,-n}^{-})-1]+\mathfrak{z}_{i,n}^{-}[\ln(\mathfrak{z}_{i,n}^{-})-1])\Big{\\}}\end{array}$
$\begin{array}[]{l}\frac{\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{-}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\,\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi)\\\ \text{and }\sin\theta_{1}-q_{2}\sin\theta_{2}=0,q^{\star}_{2}\neq 1/q_{1}^{\star}\end{array}$ | $\begin{array}[]{l}\frac{[\Gamma(z)]^{\mu_{1}^{-}-1}}{\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}\Gamma(\mathfrak{z}_{i}^{-}-z+1)}\frac{\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i}^{\star,-}-z+1)}{\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i}^{\star,-}+z)}\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{-})}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{-}+1)}\\\ \times\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,-}-z+1)}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,-}+z)}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,-}-1}(\mathfrak{z}_{i,n}^{\star,-})^{2z-1}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}[\mathfrak{z}_{i,n}^{-}(-\mathfrak{z}_{i,-n}^{-})]^{\frac{2z-1}{2}}}\\\ \times\exp\Big{\\{}\sum\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i,-n}^{-}[\ln(-\mathfrak{z}_{i,-n}^{-})-1]+\mathfrak{z}_{i,n}^{-}[\ln(\mathfrak{z}_{i,n}^{-})-1])\Big{\\}}\end{array}$
$\begin{array}[]{l}\frac{\mathcal{S}^{-}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{-}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\,\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}-q_{2}\sin\theta_{2}=0,\,q^{\star}_{2}=1/q_{1}^{\star}\end{array}$ | $\begin{array}[]{l}\frac{[\Gamma(z)]^{\mu_{1}^{-}-3}}{\prod\limits_{i=\mu_{1}^{-}+1}^{{\boldsymbol{K}}^{-}-1}\Gamma(\mathfrak{z}_{i}^{-}-z+1)}\frac{\prod\limits_{i=4}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i}^{\star,-}-z+1)}{\prod\limits_{i=4}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i}^{\star,-}+z)}\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{-})}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{-}+1)}\\\ \times\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,-}-z+1)}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,-}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,-}+z)}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,-}-1}(\mathfrak{z}_{i,n}^{\star,-})^{2z-1}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}[\mathfrak{z}_{i,n}^{-}(-\mathfrak{z}_{i,-n}^{-})]^{\frac{2z-1}{2}}}\\\ \times\exp\Big{\\{}\sum\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}(\mathfrak{z}_{i,-n}^{-}[\ln(-\mathfrak{z}_{i,-n}^{-})-1]+\mathfrak{z}_{i,n}^{-}[\ln(\mathfrak{z}_{i,n}^{-})-1])\Big{\\}}\end{array}$
$\begin{array}[]{l}\frac{\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\,\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}+q_{2}\sin\theta_{2}\neq 0\end{array}$ | $\begin{array}[]{l}\frac{\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i}^{\star,+}-z+1)}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(\mathfrak{z}_{i}^{+}-z+1)\Gamma(z)\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i}^{\star,+}+z)}\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{+})}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{+}+1)}\\\ \times\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,+}-z+1)}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,+}+z)}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}(\mathfrak{z}_{i,n}^{\star,+})^{2z-1}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}[\mathfrak{z}_{i,n}^{+}(-\mathfrak{z}_{i,-n}^{+})]^{\frac{2z-1}{2}}}\\\ \times\exp\Big{\\{}\sum\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i,-n}^{+}[\ln(-\mathfrak{z}_{i,-n}^{+})-1]+\mathfrak{z}_{i,n}^{+}[\ln(\mathfrak{z}_{i,n}^{+})-1])\Big{\\}}\end{array}$
$\begin{array}[]{l}\frac{\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1}^{\star},q_{2}^{\star})},\,\text{ if }\theta_{1},\,\theta_{2}\in(0,\pi),\\\ \sin\theta_{1}+q_{2}\sin\theta_{2}=0\end{array}$ | $\begin{array}[]{l}\frac{[\Gamma(z)]^{\mu_{1}^{+}-1}}{\prod\limits_{i=\mu_{1}^{+}+1}^{{\boldsymbol{K}}^{+}-1}\Gamma(\mathfrak{z}_{i}^{+}-z+1)}\frac{\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i}^{\star,+}-z+1)}{\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i}^{\star,+}+z)}\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{+})}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(z-\mathfrak{z}_{i,-n}^{+}+1)}\\\ \times\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,+}-z+1)}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i,n}^{\star,+}+z)}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}(\mathfrak{z}_{i,n}^{\star,+})^{2z-1}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}[\mathfrak{z}_{i,n}^{+}(-\mathfrak{z}_{i,-n}^{+})]^{\frac{2z-1}{2}}}\\\ \times\exp\Big{\\{}\sum\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i,-n}^{+}[\ln(-\mathfrak{z}_{i,-n}^{+})-1]+\mathfrak{z}_{i,n}^{+}[\ln(\mathfrak{z}_{i,n}^{+})-1])\Big{\\}}\end{array}$

It is worth noting that, exploiting the same arguments allows one to obtain
the representation of $\mathbb{L}^{\star}(z)$ and $\mathfrak{d}_{0}^{\star}$
for the rest coefficients $\mathcal{S}(z)$ described in Propositions 5.1-5.3.

Summarizing, in order to obtain an explicit solution of homogenous equation
(5.1) in the case of the coefficient $\mathcal{S}$ listed in Table
LABEL:tab:table2, we are left to substitute the corresponding values of
$\mathbb{L}^{\star}(z)$ and $\mathfrak{d}_{0}^{\star}$ from Tables
LABEL:tab:table2-LABEL:tab:table3 to expression (5.13) with $B^{\star}=0$.

######  Remark 5.1.

Appealing to the well-known relations: $\sinh(q_{0}z)=-i\sin(iq_{0}z),$
$\cosh(q_{0}z)=\cos(iq_{0}z),$ and performing the inessential modifications in
the arguments above, we obtain the very same results for homogenous equation
(5.1) with $\mathcal{S}(z)=\sinh(q_{0}z),$ $\cosh(q_{0}z)$, $\tanh(q_{0}z)$,
$\coth(q_{0}z)$ and their linear expressions.

At this point, we analyze inhomogeneous equation (5.1) with
$\mathcal{S}(z)=\frac{\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1},q_{2}^{\star})}$
and demonstrate particular solution (5.15) with different kinds of
${\mathbb{P}}_{1}(z)$.

######  Example 5.2.

We consider (5.1) with
$\mathcal{S}(z)=\frac{\mathcal{S}^{+}(z;\theta_{1},\theta_{2},q_{1},q_{2})}{\mathcal{S}^{+}(z;0,0,q_{1},q_{2}^{\star})}$
where $\theta_{1},\theta_{2}\in(0,\pi)$ and
$\sin\theta_{1}-q_{2}\sin\theta_{2}\neq 0.$ In this example, we analyze three
kinds of ${\mathbb{P}}_{1}(z)$:

$\displaystyle{\mathbb{P}}_{I}$
$\displaystyle:={\mathbb{P}}_{1}(z)=1,\quad{\mathbb{P}}_{II}:={\mathbb{P}}_{1}(z)=\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\sin\pi(\mathfrak{z}_{i}^{\star,+}-z+1)\exp\\{i\pi(\mathfrak{z}_{i}^{\star,+}-z+1)\\},$
$\displaystyle{\mathbb{P}}_{III}$
$\displaystyle:={\mathbb{P}}_{1}(z)=\frac{\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\sin\pi(\mathfrak{z}_{i}^{\star,+}-z+1)\exp\\{i\pi(\mathfrak{z}_{i}^{\star,+}-z+1)\\}}{\prod\limits_{i=1}^{{\boldsymbol{K}}^{+}-1}\sin\pi(\mathfrak{z}_{i}^{+}-z+1)\exp\\{i\pi(\mathfrak{z}_{i}^{+}-z+1)+i\pi
z\\}\sin\pi z}$

and two types of $\mathbb{F}(z,\sigma)$. Namely, the main difference of these
kinds is related with the behavior of $\mathbb{F}(z,\sigma)$ as $|Imz|$ tends
to the infinite for each fixed $Rez$ and $\sigma\in{\mathbb{C}}$,
$Re\sigma>0.$

Indeed, we assume that $\mathbb{F}$ does not have any poles for
$z\in{\mathbb{C}}$ and $\sigma\in{\mathbb{C}}$ with $Re\sigma>0$ and

i: either

$\underset{|Imz|\to+\infty}{\lim}\,|F(z,\sigma)|=C,$ (5.16)

ii: or

$\underset{|Imz|\to+\infty}{\lim}\,e^{c^{\star}\pi|Imz|}|F(z,\sigma)|=C,$
(5.17)

with $c^{\star}=\max\\{4,2{\boldsymbol{K}}^{\star,+}\\}$.

It is apparent that the functions $\mathbb{F}$ considered in this example meet
the requirements of Theorem 5.2. It is worth noting that, assumptions H12-H13
tell us that if for each $m\in\mathbb{N}\cup\\{0\\}$ the relations hold

$Rez\neq
m+\mathfrak{z}_{i,n}^{\star,+}+1,\,i\in\\{0,...,{\boldsymbol{K}}^{\star,+}-1\\},\,n\in\mathbb{N}\cup\\{0\\};\,Rez\neq\mathfrak{z}_{i,-n}^{+}-m,\,i\in\\{0,...,{\boldsymbol{K}}^{+}-1\\},\,n\in\mathbb{N},$
(5.18)

then the function
$\frac{Y_{h}(z,\sigma;{\mathbb{P}})}{{\mathbb{P}}(z,\sigma)}$ has no poles,
while
$\frac{{\mathbb{P}}(z+1+\xi,\sigma)}{Y_{h}(z+1+\xi,\sigma;{\mathbb{P}})}$ has
no poles if for $m\in\mathbb{N}\cup\\{0\\}$ the system is fulfilled

$\begin{cases}Rez\neq m+d_{0}+\mathfrak{z}_{i,n}^{+},\quad
i\in\\{0,1,...,{\boldsymbol{K}}^{+}-1\\},\,n\in\mathbb{N}\cup\\{0\\},\\\
Rez\neq-m-1+d_{0},\\\ Rez\neq-m+d_{0}-1-\mathfrak{z}_{i,n}^{\star,+},\quad
i\in\\{0,1,...,{\boldsymbol{K}}^{\star,+}-1\\},\,n\in\mathbb{N},\\\
Rez\neq-m+d_{0}-\mathfrak{z}_{i}^{+},\quad
i\in\\{1,...,{\boldsymbol{K}}^{\star,+}-1\\}.\\\ \end{cases}$ (5.19)

Nevertheless, the function ${\mathbb{P}}(z)$ of the second and the third kinds
permit to relax conditions (5.18) and (5.19). Indeed, in the case of
${\mathbb{P}}:={\mathbb{P}}_{II}$, this function is chosen so as to eliminate
the poles of the functions $\Gamma(\mathfrak{z}_{i}^{\star,+}-z+1),$
$i\in\\{1,...,{\boldsymbol{K}}^{\star,+}-1\\}$, which means the first
inequality in (5.18) should be satisfied only for $n\in\mathbb{N}$. Thus, we
can extend the domain of analyticity to the homogenous solution
$Y_{h}(z,\sigma;{\mathbb{P}}_{II})$. However, the such choice of
${\mathbb{P}}(z)$ provides additional simple nulls to
$Y_{h}(z,\sigma;{\mathbb{P}}_{II})$ or in turn, additional poles of
$\frac{1}{Y_{h}(z+1+\xi,\sigma;{\mathbb{P}}_{II})}$. To avoid this problem, we
are left to require that $Rez\neq\mathfrak{z}_{i}^{\star,+}+d_{0}-m,\quad
m\in\mathbb{N}.$ Coming to the function ${\mathbb{P}}_{III}(z)$, clearly, the
function
$\frac{{\mathbb{P}}_{III}(z)\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i}^{\star,+}-z+1)}{\Gamma(z)\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(\mathfrak{z}_{i}^{+}-z+1)}$
does not have any poles if

$Rez\neq\mathfrak{z}_{i}^{+}-m+1,\quad
i\in\\{0,...,{\boldsymbol{K}}^{+}-1\\},\quad\text{and}\quad Rez\neq m,$

while the function
$\frac{\Gamma(z+\xi+1)\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(\mathfrak{z}_{i}^{+}-z-\xi)}{{\mathbb{P}}_{III}(z+1+\xi)\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\Gamma(\mathfrak{z}_{i}^{\star,+}-z-\xi)}$
does not have any poles if

$Rez\neq-m+\mathfrak{z}_{i}^{\star,+}+d_{0},\quad
i\in\\{1,2,...,{\boldsymbol{K}}^{\star,+}-1\\}.$

Thus, this choice of ${\mathbb{P}}(z)$ allows us to eliminate the poles of
$\Gamma(\mathfrak{z}_{i}^{\star,+}-z+1)$,
$i\in\\{1,2,...,{\boldsymbol{K}}^{\star,+}-1\\}$ in the case of the function
$Y_{h}(z,\sigma;{\mathbb{P}}_{III})$ and to remove the poles of
$\Gamma(z+\xi+1)$ and $\Gamma(\mathfrak{z}_{i}^{+}-z-\xi),$
$i\in\\{0,1,...,{\boldsymbol{K}}^{+}-1\\}$ in the case of the function
$\frac{1}{Y_{h}(z+1+\xi,\sigma;{\mathbb{P}}_{III})}$.

In summary, to construct the solution of inhomogeneous equation (5.1), we are
left to select the suitable kernel $\mathcal{K}_{1}(\xi)$ such that equality
(5.14) holds. Appealing to Theorem 5.1 and asymptotic (2.5) and performing
technical calculations, we end up with inequalities:

$\displaystyle\Big{|}\frac{e^{-(\varepsilon+\theta_{2})|Im\,z|}}{Y_{h}(z,\sigma;{\mathbb{P}}_{I})}\Big{|}\leq
C\begin{cases}e^{\frac{3\pi}{2}Im\,z},\,Im\,z\to+\infty,\\\
e^{\frac{\pi}{2}Im\,z},\,Im\,z\to-\infty,\end{cases}\,\Big{|}\frac{e^{-(\varepsilon+\theta_{2})|Im\,z|}}{Y_{h}(z,\sigma;{\mathbb{P}}_{II})}\Big{|}\leq
C\begin{cases}e^{\frac{3\pi}{2}Im\,z},\qquad\quad\quad\,Im\,z\to+\infty,\\\
e^{-2\pi[{\boldsymbol{K}}^{\star,+}-\frac{5}{4}]Im\,z},\,Im\,z\to-\infty,\end{cases}$
$\displaystyle\Big{|}\frac{e^{-(\varepsilon+\theta_{2})|Im\,z|}}{Y_{h}(z,\sigma;{\mathbb{P}}_{III})}\Big{|}\leq
C\begin{cases}e^{(\frac{11\pi}{2}-2\pi{\boldsymbol{K}}^{\star,+})Im\,z},\,Im\,z\to+\infty,\\\
e^{-\frac{3\pi}{2}Im\,z},\qquad\qquad Im\,z\to-\infty,\end{cases}$

where $\varepsilon>0$ is enough small value, e.g.
$\varepsilon<<\frac{\pi}{100}$. Thus, taking into account these relations and
behavior of the function $|\mathbb{F}|$ as $|Imz|\to+\infty$ (see (5.16) and
(5.17)), we arrive at
$\mathcal{K}_{1}(\xi):=\mathcal{K}_{1}(\xi;{\mathbb{P}}_{j})$, $j=I,II,III$
listed in Tables LABEL:tab:table4-LABEL:tab:table5. It is apparent that, decay
of $|\mathbb{F}(z,\sigma)|$ as $|Imz|\to+\infty$ permits to relax assumption
on $\mathcal{K}_{1}(\xi,{\mathbb{P}}_{j}),$ $j=I,II,III$.

Table 4. The value of $\mathcal{K}_{1}(\xi)$ in the case of $\mathbb{F}(z,\sigma)$ satisfying (5.17) $Y_{h}(z,\sigma;{\mathbb{P}}_{j})$ | $\mathcal{K}_{1}(\xi,{\mathbb{P}}_{j}),\,j=I,II,III$
---|---
$Y_{h}(z,\sigma;{\mathbb{P}}_{I}),Y_{h}(z,\sigma;{\mathbb{P}}_{II})$ | $C$
$Y_{h}(z,\sigma;{\mathbb{P}}_{III})$ | $\sin^{2}\pi d\,\sin^{-2}\pi(\xi+d)$
Table 5. The value of $\mathcal{K}_{1}(\xi)$ in the case of $\mathbb{F}(z,\sigma)$ satisfying (5.16) $Y_{h}(z,\sigma;{\mathbb{P}}_{j})$ | $\mathcal{K}_{1}(\xi,{\mathbb{P}}_{j}),\,j=I,II,III$
---|---
$Y_{h}(z,\sigma;{\mathbb{P}}_{I})$ | $\sin^{2}\pi d\sin^{-2}\pi(\xi+d)$
$Y_{h}(z,\sigma;{\mathbb{P}}_{II})$ if $\pi(\frac{5}{2}-2{\boldsymbol{K}}^{\star,+})-\theta_{2}-\varepsilon>0$ | $\sin^{2}\pi d\sin^{-2}\pi(\xi+d)$
$Y_{h}(z,\sigma;{\mathbb{P}}_{II})$ if $\pi(\frac{5}{2}-2{\boldsymbol{K}}^{\star,+})-\theta_{2}-\varepsilon<0$ | $(\sin\pi d)^{2{\boldsymbol{K}}^{\star,+}}(\sin\pi(\xi+d))^{-2{\boldsymbol{K}}^{\star,+}}$
$Y_{h}(z,\sigma;{\mathbb{P}}_{III})$ | $\sin^{4}\pi d\sin^{-4}\pi(\xi+d)$

Finally, collecting the results above with Theorems 5.1-5.2, we derive the
general solution to (5.1) in the form

$Y(z,\sigma)=Y_{h}(z,\sigma;{\mathbb{P}})+Y_{h}(z,\sigma;{\mathbb{P}}_{j})\int\limits_{\ell_{d_{0}}}\frac{\mathbb{F}(z+\xi,\sigma)[\cot\pi\xi+i]\mathcal{K}_{1}(\xi,{\mathbb{P}}_{j})}{[a_{1}\sigma+a_{2}\sigma^{\nu}]Y_{h}(z+\xi+1,\sigma;{\mathbb{P}}_{j})}d\xi,\,j=I,II,III,$
(5.20)

where $\mathcal{K}_{1}(\xi;{\mathbb{P}}_{j})$ are given in Tables
LABEL:tab:table5-LABEL:tab:table4, and

$\displaystyle Y_{h}(z,\sigma;{\mathbb{P}})$
$\displaystyle=\bigg{[}\frac{(\sin\theta_{1}+q_{2}\sin\theta_{2})\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}(\mathfrak{z}_{i}^{\star,+})^{2}}{(1+q_{2}^{\star}q_{1})\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\mathfrak{z}_{i}^{+}}\bigg{]}^{z-1/2}\frac{e^{i\pi
z}(a_{1}\sigma+a_{2}\sigma^{\nu})^{\frac{1-2z}{2}}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Gamma(z)\Gamma(\mathfrak{z}_{i}^{+}-z+1)}$
$\displaystyle\times\prod\limits_{i=1}^{{\boldsymbol{K}}^{\star,+}-1}\frac{\Gamma(\mathfrak{z}_{i}^{\star,+}-z+1)}{\Gamma(\mathfrak{z}_{i}^{\star,+}+z)}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\frac{\Gamma(z-\mathfrak{z}_{i,-n}^{+})}{\Gamma(\mathfrak{z}_{i,n}^{+}-z+1)}\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}\frac{\Gamma(\mathfrak{z}_{i,n}^{\star,+}-z+1)}{\Gamma(\mathfrak{z}_{i,n}^{\star,+}+z)}$
$\displaystyle\times\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{\star,+}-1}(\mathfrak{z}_{i,n}^{\star,+})^{2z-1}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}[\mathfrak{z}_{i,n}^{+}(-\mathfrak{z}_{i,-n}^{+})]^{\frac{2z-1}{2}}}\exp\Big{\\{}\sum\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}(\mathfrak{z}_{i,-n}^{+}[\ln(-\mathfrak{z}_{i,-n}^{+})-1]+\mathfrak{z}_{i,n}^{+}[\ln(\mathfrak{z}_{i,n}^{+})-1])\Big{\\}}.$

It is worth noting that, if we search a bounded solution of (5.1) if
$|Imz|\to+\infty$, then we should put ${\mathbb{P}}\equiv 0$ in
$Y_{h}(z,\sigma;{\mathbb{P}})$, which means $Y_{h}(z,\sigma;{\mathbb{P}})=0$
and (5.20) is rewritten as $Y(z,\sigma)=Y_{ih}(z,\sigma).$

######  Remark 5.2.

Clearly, $\mathbb{F}(z,\sigma)=1$ meets requirements of Theorem 5.2 and
satisfies equality (5.16), while
$\mathbb{F}(z,\sigma)=\exp\\{-c^{\star}(z\bar{z})\\}$ for each fixed $Rez$
satisfies equality (5.17).

######  Remark 5.3.

Observing arguments of Sections 3-5, one can easily conclude that all results
of Section 5 (possibly with minor modifications) can be extended to the
functions $\mathcal{S}$ presented as either
$\mathcal{S}=\prod_{j=1}^{N_{I}}\mathcal{S}_{1}^{j}$ or
$\mathcal{S}=\frac{\prod_{j=1}^{N_{I}}\mathcal{S}_{1}^{j}}{\prod_{j=1}^{N_{II}}\mathcal{S}_{2}^{j}}$
with fixed integer $N_{I}$ and $N_{II}$ and the functions
$\mathcal{S}_{1}^{j}$ and $\mathcal{S}_{2}^{j}$ having the properties of the
functions $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$, respectively.

Besides, in light of Remark 2.5, the results of Theorems 5.1 and 5.2 hold for
multidimensional equation similar to (2.11) with
$\mathbf{S}(z)=\prod_{j=1}^{k}\mathcal{S}^{j}(z)$, where $\mathcal{S}^{j}(z)$
obeys properties of the function $\mathcal{S}(z)$.

It is worth noting that, results obtained in Section 5 mean that FDEs studied
in [4, 5, 6, 7, 11, 22, 23, 27] (see also references therein) fall in our
analysis.

##  6\. Explicit solutions of transmission boundary value problems with
a fractional dynamic boundary condition in plane corners

In this Section, we discuss a construction of a solution to a transmission
problem (1.5)-(1.9) via results of Sections 2 and 5. Other words, we
demonstrate the application of solutions (1.1) to look for an explicit
solution to the non-classical transmission boundary value problem.

We assume that $\omega_{0}=q\pi/p\in[0,\pi/2)$ with $p$ and $q$ satisfying
assumptions of Corollary 5.1. Coming to the coefficients involved, $a_{1}$ and
$a_{2}$ meet requirements H1, and $a_{3}$, $a_{4},$ $\mathfrak{K}\neq 1$ are
positive quantities, and $\mathfrak{s}_{0}$ is a given real number.

As for the right-hand sides in (1.5)-(1.9), we state the following hypothesis.

H14: We require that for some fixed $\alpha\in(0,1),$

$f_{1},f\in\mathcal{C}([0,T],\mathcal{C}^{1+\alpha}(g)),\,\,f_{2}\in\mathcal{C}([0,T],\mathcal{C}^{\alpha}(\partial
G_{1}\backslash
g)),\,\,f_{3}\in\mathcal{C}([0,T],\mathcal{C}^{\alpha}(\partial
G_{2}\backslash g)),$
$f_{0,1}\in\mathcal{C}([0,T],\mathcal{C}^{\alpha}(\bar{G}_{1})),\,f_{0,2}\in\mathcal{C}([0,T],\mathcal{C}^{\alpha}(\bar{G}_{2})),$

and

$f,f_{1},f_{2},f_{3},f_{0,1},f_{0,2}\equiv 0\quad\text{if either}\quad
r\in[0,\varepsilon_{1}]\cup[R_{1},+\infty)\quad\text{or}\quad t\leq 0,$

for some fixed small positive $\varepsilon_{1}$ and enough large $R_{1}$.

It is apparent that assumptions on the smoothness of the right-hand sides may
not be enough to prove existence of a classical smooth solution, but H14 is
enough to construct explicit solutions to (1.5)-(1.9). Here we focus on the
solution of the problem which vanishes as $r\to+\infty$. First of all, we
remark that the explicit solutions of classical Dirichlet-transmission problem
to Poisson equations in domain $G_{1}\cup g\cup G_{2}$ were searched in
Section 9.4 [32]. In light of this fact, we are left to construct solutions in
the case of homogenous equations (1.5) and homogenous Dirichlet and
transmission conditions (1.8) and (1.9), i.e. we put

$f_{0,1},f_{0,2},f_{1},f_{2},f_{3}\equiv 0.$ (6.1)

In order to build the solution of (1.5)-(1.9), we will exploit the following
strategy. In the first stage, employing special change of variables and then
Fourier and Laplace transforms, we reduce problem (1.5)-(1.9) to a functional
difference equation. The second stage is related with applying Theorems 2.1
and 2.2 to this equation, and in turn with obtaining solution via formula like
(2.4), (2.9) and (2.10). Finally, performing the inverse Laplace and Fourier
transformation, we conclude with the explicit integral form of the solution to
(1.5)-(1.9).

$\bullet$ The First Stage. We start with standard change of variable (see e.g.
(2.6) in [4] or (3.9) in [5]) in the case of a boundary value problem stated
in the domain with corner points:

$x_{1}=\ln r,\quad x_{2}=\varphi.$ (6.2)

This map transforms the corners $G_{1}$ and $G_{2}$ to the strips
$\mathfrak{G}_{1}$ and $\mathfrak{G}_{2}$, respectively,

$\mathfrak{G}_{1}=\\{(x_{1},x_{2}):\,x_{1}\in\mathbb{R},\,-\pi/2<x_{2}<\omega_{0}\\},\quad\mathfrak{G}_{2}=\\{(x_{1},x_{2}):\,x_{1}\in\mathbb{R},\,\omega_{0}<x_{2}<\pi/2\\},$

while the image of the boundary $g$ is
$\mathfrak{g}=\\{(x_{1},x_{2}):\,x_{1}\in\mathbb{R},\,x_{2}=\omega_{0}\\}.$

Performing the change of variables (6.2) in relations (1.5)-(1.9) and using
the same notations for the functions $u_{1}$ and $u_{2}$ and $f$ in the new
variables, we end up with the problem

$\begin{cases}\Delta
u_{1}=0\qquad\text{in}\qquad\mathfrak{G}_{1,T},\qquad\quad
u_{1}(x_{1},x_{2},0)=0\qquad\text{in}\qquad\bar{\mathfrak{G}}_{1},\\\ \Delta
u_{2}=0\qquad\text{in}\qquad\mathfrak{G}_{2,T},\qquad\quad
u_{2}(x_{1},x_{2},0)=0\qquad\text{in}\qquad\bar{\mathfrak{G}}_{2},\\\
\exp\\{(\mathfrak{s}_{0}+1)x_{1}\\}\Big{[}a_{1}\frac{\partial}{\partial
t}(u_{1}-u_{2})+a_{2}\mathbb{D}_{t}^{\nu}(u_{1}-u_{2})\Big{]}\\\
-\frac{\partial}{\partial x_{2}}(u_{1}-u_{2})+a_{3}\frac{\partial}{\partial
x_{1}}(u_{1}-u_{2})=e^{x_{1}}f(x_{1},t)\qquad\quad\text{on}\qquad\mathfrak{g}_{T},\\\
\frac{\partial u_{1}}{\partial x_{2}}-\mathfrak{K}\frac{\partial
u_{2}}{\partial x_{2}}+\mathfrak{K}a_{4}\frac{\partial}{\partial
x_{1}}(u_{1}-u_{2})=0\qquad\qquad\qquad\qquad\text{on}\quad\mathfrak{g}_{T},\\\
u_{1}(x_{1},-\pi/2,t)=0\quad\text{and}\quad
u_{2}(x_{1},\pi/2,t)=0\qquad\qquad\text{for}\quad
x_{1}\in\mathbb{R},\,t\in[0,T].\end{cases}$ (6.3)

After that, as usual to the analysis of boundary value problems in domains
with singular boundaries (see e.g. [16, Section 4]), we will search a solution
to (6.3) in the form:

$u_{1}(x_{1},x_{2},t)=e^{\mathfrak{s}x_{1}}U_{1}(x,_{1},x_{2},t),\quad
u_{2}(x_{1},x_{2},t)=e^{\mathfrak{s}x_{1}}U_{2}(x,_{1},x_{2},t),$

where the value $\mathfrak{s}\neq 0$ will be specified below, and new unknown
functions solve the problem

$\begin{cases}\Delta U_{1}+2\mathfrak{s}\frac{\partial U_{1}}{\partial
x_{1}}+\mathfrak{s}^{2}U_{1}=0\qquad\text{in}\quad\mathfrak{G}_{1,T},\qquad\quad
U_{1}(x_{1},x_{2},0)=0\qquad\text{in}\quad\bar{\mathfrak{G}}_{1},\\\ \Delta
U_{2}+2\mathfrak{s}\frac{\partial U_{2}}{\partial
x_{1}}+\mathfrak{s}^{2}U_{2}=0\qquad\text{in}\quad\mathfrak{G}_{2,T},\quad\qquad
U_{2}(x_{1},x_{2},0)=0\qquad\text{in}\quad\bar{\mathfrak{G}}_{2},\\\
\exp\\{(\mathfrak{s}_{0}+1)x_{1}\\}\Big{[}a_{1}\frac{\partial}{\partial
t}(U_{1}-U_{2})+a_{2}\mathbb{D}_{t}^{\nu}(U_{1}-U_{2})\Big{]}\\\
-\frac{\partial}{\partial
x_{2}}(U_{1}-U_{2})+a_{3}\Big{[}\frac{\partial}{\partial
x_{1}}(U_{1}-U_{2})+\mathfrak{s}(U_{1}-U_{2})\Big{]}=e^{(1-\mathfrak{s})x_{1}}f(x_{1},t)\quad\text{on}\quad\mathfrak{g}_{T},\\\
\frac{\partial U_{1}}{\partial x_{2}}-\mathfrak{K}\frac{\partial
U_{2}}{\partial x_{2}}+\mathfrak{K}a_{4}\Big{[}\frac{\partial}{\partial
x_{1}}(U_{1}-U_{2})+\mathfrak{s}(U_{1}-U_{2})\Big{]}=0\qquad\qquad\qquad\qquad\text{on}\quad\mathfrak{g}_{T},\\\
U_{1}(x_{1},-\pi/2,t)=0\qquad\text{and}\qquad
U_{2}(x_{1},\pi/2,t)=0\quad\qquad\qquad\text{for}\quad
x_{1}\in\mathbb{R},\,t\in[0,T].\end{cases}$

It is worth noting that the introducing new functions $U_{1}$ and $U_{2}$ and
rewriting problem (6.3) in the terms of these functions permit to study the
corresponding problem in either Hölder or Sobolev spaces instead of using
weighted spaces (see for detail e.g. [4, Section2] or [7, Section 3.1]).

Then, we denote by $\tilde{U}(\lambda,x_{2},t)$ the Fourier transform of
$U(x_{1},x_{2},t)$, and by $\hat{U}(x_{1},x_{2},\sigma)$ the Laplace transform
of $U(x_{1},x_{2},t)$, and use the notation $*$ instead of $\hat{\tilde{\,}}$.
Thus, performing these transformations in the relations above, and setting
$\mathfrak{x}=i\lambda+\mathfrak{s},$
$\mathfrak{s}^{\star}=\mathfrak{s}_{0}+1,$ we have

$\begin{cases}\frac{\partial^{2}U^{*}_{1}}{\partial
x_{2}^{2}}+\mathfrak{x}^{2}U^{*}_{1}=0,\quad\tilde{U_{1}}(\lambda,x_{2},0)=0,\quad
x_{2}\in(-\pi/2,\omega_{0}),\\\ \frac{\partial^{2}U^{*}_{2}}{\partial
x_{2}^{2}}+\mathfrak{x}^{2}U^{*}_{2}=0,\quad\tilde{U_{2}}(\lambda,x_{2},0)=0,\quad
x_{2}\in(\omega_{0},\pi/2),\\\
(a_{1}\sigma+a_{2}\sigma^{\nu})[U_{1}^{*}(\lambda+i\mathfrak{s}^{\star},\omega_{0},\sigma)-U_{2}^{*}(\lambda+i\mathfrak{s}^{\star},\omega_{0},\sigma)]-\frac{\partial}{\partial
x_{2}}[U_{1}^{*}(\lambda,\omega_{0},\sigma)-U_{2}^{*}(\lambda,\omega_{0},\sigma)]\\\
+a_{3}\mathfrak{x}[U_{1}^{*}(\lambda,\omega_{0},\sigma)-U_{2}^{*}(\lambda,\omega_{0},\sigma)]=f^{*}(\lambda+i(1-\mathfrak{s}),\sigma),\\\
\frac{\partial U_{1}^{*}}{\partial
x_{2}}(\lambda,\omega_{0},\sigma)-\mathfrak{K}\frac{\partial
U_{2}^{*}}{\partial
x_{2}}(\lambda,\omega_{0},\sigma)+\mathfrak{K}a_{4}\mathfrak{x}[U_{1}^{*}(\lambda,\omega_{0},\sigma)-U_{2}^{*}(\lambda,\omega_{0},\sigma)]=0,\\\
U_{1}^{*}(\lambda,-\pi/2,\sigma)=0,\quad
U_{2}^{*}(\lambda,\pi/2,\sigma)=0.\end{cases}$ (6.4)

To satisfy equations and boundary conditions on $x_{2}=\pm\pi/2$, we set

$U_{1}^{\star}(\lambda,x_{2},\sigma)=\mathcal{M}_{1}(\lambda,\sigma)\sin\mathfrak{x}(x_{2}+\pi/2),\quad
U_{2}^{\star}(\lambda,x_{2},\sigma)=\mathcal{M}_{2}(\lambda,\sigma)\sin\mathfrak{x}(x_{2}-\pi/2),$
(6.5)

where unknown functions $\mathcal{M}_{1}(\lambda,\sigma)$ and
$\mathcal{M}_{2}(\lambda,\sigma)$ will be found via transmission conditions in
(6.4). To this end, substituting $U_{1}^{*}$ and $U^{*}_{2}$ to the
transmission conditions and putting

$\mathcal{N}(\lambda)=\frac{\cos\mathfrak{x}(\omega_{0}-\pi/2)+a_{4}\sin\mathfrak{x}(\omega_{0}-\pi/2)}{\cos\mathfrak{x}(\omega_{0}+\pi/2)+\mathfrak{K}a_{4}\sin\mathfrak{x}(\omega_{0}+\pi/2)},$

we arrive at the system with unknown functions $\mathcal{M}_{1}$ and
$\mathcal{M}_{2}$:

$\begin{cases}\mathcal{M}_{1}(\lambda,\sigma)=\mathfrak{K}\mathcal{N}(\lambda)\mathcal{M}_{2}(\lambda,\sigma),\\\
[a_{1}\sigma+a_{2}\sigma^{\nu}]\mathcal{M}_{2}(\lambda+i\mathfrak{s}^{\star},\sigma)[\mathfrak{K}\mathcal{N}(\lambda+i\mathfrak{s}^{\star})\sin\mathfrak{x}(\omega_{0}+\pi/2)-\sin\mathfrak{x}(\omega_{0}-\pi/2)]\\\
-\mathfrak{x}\mathcal{M}_{2}(\lambda,\sigma)\\{a_{3}\sin\mathfrak{x}(\omega_{0}-\pi/2)-\cos\mathfrak{x}(\omega_{0}-\pi/2)\\\
+\mathfrak{K}\mathcal{N}(\lambda)[\cos\mathfrak{x}(\omega_{0}+\pi/2)-a_{3}\sin\mathfrak{x}(\omega_{0}-\pi/2)]\\}=f^{*}(\lambda+i(1-\mathfrak{s}),\sigma).\end{cases}$
(6.6)

In virtue of this system, we are left to look for the function
$\mathcal{M}_{2}$. Denoting

$\mathcal{N}_{1}(\lambda)=\mathfrak{K}\mathcal{N}\sin\mathfrak{x}(\omega_{0}+\pi/2)-\sin\mathfrak{x}(\omega_{0}-\pi/2)$

and substituting the expression of $\mathcal{M}_{1}$ to the second equation in
(6.6), we obtain

$\displaystyle\mathfrak{x}\mathcal{M}_{2}(\lambda,\sigma)\\{\mathfrak{K}\mathcal{N}(\lambda)\cos\mathfrak{x}(\omega_{0}+\pi/2)-a_{3}\mathcal{N}_{1}(\lambda)-\cos\mathfrak{x}(\omega_{0}-\pi/2)\\}$
$\displaystyle-[a_{1}\sigma+a_{2}\sigma^{\nu}]\mathcal{M}_{2}(\lambda+i\mathfrak{s}^{\star},\sigma)\mathcal{N}_{1}(\lambda+i\mathfrak{s}^{\star})=-f^{*}(\lambda+i(1-\mathfrak{s}),\sigma)$

which in turn is reduced to a functional equation in unknown function
$\mathcal{U}:=\mathcal{M}_{2}(\lambda,\sigma)\mathcal{N}_{1}(\lambda),$

$[a_{1}\sigma+a_{2}\sigma^{\nu}]\mathcal{U}(\lambda+i\mathfrak{s}^{\star},\sigma)-\mathfrak{x}\mathcal{G}(\lambda)\mathcal{U}(\lambda,\sigma)=f^{*}(\lambda+i(1-\mathfrak{s}),\sigma).$

Here we put

$\mathcal{G}(\lambda)=\frac{\mathfrak{K}\mathcal{N}(\lambda)\cos\mathfrak{x}(\omega_{0}+\pi/2)-\cos\mathfrak{x}(\omega_{0}-\pi/2)}{\mathcal{N}_{1}(\lambda)}-a_{3}.$

######  Remark 6.1.

It is apparent that if $\mathfrak{s}_{0}=-1,$ i.e. $\mathfrak{s}^{\star}=0$,
then the function
$\mathcal{U}=\frac{f^{*}(\lambda+i(1-\mathfrak{s}),\sigma)}{a_{1}\sigma+a_{2}\sigma^{\nu}-\mathfrak{x}\,\mathcal{G}(\lambda)}$
solves the equation above, which in turn entails

$U^{*}_{1}=\mathfrak{K}\sin\mathfrak{x}(x_{2}+\pi/2)\mathcal{U}(\lambda,\sigma)\mathcal{N}(\lambda)\mathcal{N}_{1}(\lambda),\qquad
U^{*}_{2}=\sin\mathfrak{x}(x_{2}-\pi/2)\mathcal{U}(\lambda,\sigma)\mathcal{N}_{1}(\lambda).$

Taking into account this remark, we will assume further
$\mathfrak{s}^{\star}\neq 0$ and introduce the new variable

$\lambda=i\mathfrak{s}^{\star}\rho,$ (6.7)

and new unknown function
$V(\rho,\sigma)=\mathcal{U}(i\mathfrak{s}^{\star}\rho,\sigma),$ we rewrite the
equation above in the form

$[a_{1}\sigma+a_{2}\sigma^{\nu}]V(\rho+1,\sigma)-[\mathfrak{s}-\mathfrak{s}^{\star}\rho]\mathcal{G}(i\mathfrak{s}^{\star}\rho)V(\rho,\sigma)=\mathfrak{F}(\rho,\sigma),$
(6.8)

where
$\mathfrak{F}(\rho,\sigma):=f^{*}(i\mathfrak{s}^{\star}\rho+i(1-\mathfrak{s}),\sigma)$.

Thus, we have reduced original problem (1.5)-(1.9) to functional equation
(6.8).

The Second Stage. In order to apply the results of Sections 2 and 5 to (6.8),
we are left to factorize the function
$\mathcal{G}(i\mathfrak{s}^{\star}\rho)$. To this end, coming to the function
$\mathcal{G}(\lambda)$, we substitute $\mathcal{N}(\lambda)$ and
$\mathcal{N}_{1}(\lambda)$ to the expression of $\mathcal{G}(\lambda)$ and
performing tedious technical calculations, we derive

$\mathcal{G}(\lambda)=\frac{\cos\pi\mathfrak{x}-\frac{a_{3}(1+\mathfrak{K}\,)+2\mathfrak{K}\,a_{4}}{\mathfrak{K}-1}\sin\pi\mathfrak{x}+\cos
2\omega_{0}\mathfrak{x}-a_{3}\sin 2\omega_{0}\mathfrak{x}}{\sin
2\omega_{0}\mathfrak{x}+\frac{\mathfrak{K}+1}{\mathfrak{K}-1}\sin\pi\mathfrak{x}}.$

Finally, denoting

$\displaystyle
q_{2}^{*}=\frac{\mathfrak{K}+1}{|\mathfrak{K}-1|},\,q_{2}=\Big{(}\frac{[a_{3}(\mathfrak{K}+1)+2\mathfrak{K}\,a_{4}]^{2}+(\mathfrak{K}-1)^{2}}{(\mathfrak{K}-1)^{2}(1+a_{3}^{2})}\Big{)}^{\frac{1}{2}},\,\,\sin\theta_{1}=(1+a_{3}^{2})^{-\frac{1}{2}},\,\cos\theta_{1}=a_{3}(1+a_{3}^{2})^{-\frac{1}{2}},$
$\displaystyle\sin\theta_{2}=\Big{(}1+\frac{[2\mathfrak{K}\,a_{4}+a_{3}(1+\mathfrak{K})]^{2}}{[\mathfrak{K}-1]^{2}}\Big{)}^{-\frac{1}{2}},\,\cos\theta_{2}=\frac{[a_{3}(\mathfrak{K}+1)+2\mathfrak{K}\,a_{4}]\text{sgn}(\mathfrak{K}-1)}{\sqrt{(\mathfrak{K}-1)^{2}+[2\mathfrak{K}\,a_{4}+a_{3}(1+\mathfrak{K})]^{2}}},$

we rewrite $\mathcal{G}(\lambda)$ in the form

$\mathcal{G}(\lambda)=-\frac{\sqrt{1+a_{3}^{2}}[\sin(2\mathfrak{x}\omega_{0}-\theta_{1})+q_{2}\sin(\pi\mathfrak{x}-\theta_{2})]}{\sin
2\omega_{0}\mathfrak{x}+\text{sgn}(\mathfrak{K}-1)q_{2}^{*}\sin\pi\mathfrak{x}}.$

Keeping in mind of assumptions on the coefficients, we easily conclude

$\theta_{1}\in(0,\pi/2)\quad\text{and}\quad\theta_{2}\in(0,\pi),\quad
q_{2}^{*}>1,\quad
q_{2}>1,\quad\sin\theta_{1}+q_{2}\sin\theta_{2}=\frac{2}{\sqrt{1+a_{3}^{2}}}.$
(6.9)

Then, performing the change of variable (6.7) and taking into account the
relations $\frac{\pi}{2\omega_{0}}=\frac{p}{2q}>1,$ we deduce

$\mathcal{G}(i\mathfrak{s}^{\star}\rho)=-\sqrt{1+a_{3}^{2}}\begin{cases}\frac{\mathcal{S}^{+}(2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho);\theta_{1},\theta_{2},\frac{p}{2q},q_{2})}{\mathcal{S}^{+}(2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho);0,0,\frac{p}{2q},q^{*}_{2})},\quad\mathfrak{K}>1,\\\
\frac{\mathcal{S}^{+}(2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho);\theta_{1},\theta_{2},\frac{p}{2q},q_{2})}{\mathcal{S}^{-}(2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho);0,0,\frac{p}{2q},q^{*}_{2})},\quad\mathfrak{K}<1,\end{cases}$

This representation means that the function
$\mathcal{G}(i\mathfrak{s}^{\star},\rho)$ is the function $\mathcal{S}$ in the
FTK case (see Section 5). Appealing to Corollary 5.1, we denote by

$\displaystyle\bar{z}_{i,n}^{+}$
$\displaystyle:=\bar{z}_{i,n}^{+}(\theta_{1},\theta_{2},q_{2}),\quad\bar{z}_{i,-n}^{+}:=\bar{z}_{i,n}^{+}(\theta_{1},\theta_{2},q_{2}),\quad\bar{{\boldsymbol{K}}}^{+}:=\bar{{\boldsymbol{K}}}^{+}(\theta_{1},\theta_{2},q_{2}),$
$\displaystyle z_{i,n}^{+}$
$\displaystyle:=z_{i,n}^{+}(0,0,q^{*}_{2}),\quad{\boldsymbol{K}}^{+}:={\boldsymbol{K}}^{+}(0,0,q^{*}_{2}),\quad
z_{i,n}^{-}:=z_{i,n}^{-}(0,0,q^{*}_{2}),\quad{\boldsymbol{K}}^{-}:={\boldsymbol{K}}^{-}(0,0,q^{*}_{2}),$

the zeros and their number in $[0,4\pi q)$ of the functions
$\mathcal{S}^{+}(z;\theta_{1},\theta_{2},\frac{p}{2q},q_{2}),$
$\mathcal{S}^{+}(z;0,0,\frac{p}{2q},q^{*}_{2})$ and $\mathcal{S}^{-}(z;0,$
$0,\frac{p}{2q},q^{*}_{2}),$ respectively. In virtue of (6.9), we can employ
Proposition 5.1 to conclude that

$\displaystyle\mathcal{G}(i\mathfrak{s}^{\star}\rho)=\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}[\bar{z}_{i}^{+}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})]\prod\limits_{i=1}^{{\boldsymbol{K}}^{+}-1}(z_{i}^{+})^{2}}{\omega_{0}(1+q_{2}^{*}q_{1})(\mathfrak{s}-\mathfrak{s}^{*}\rho)\prod\limits_{i=1}^{{\boldsymbol{K}}^{+}-1}[z_{i}^{+}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})][z_{i}^{+}-2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})]\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{z}_{i}^{+}}$
$\displaystyle\times\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\Big{[}\bar{z}_{i,n}^{+}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})\Big{]}\Big{[}-\bar{z}_{i,-n}^{+}+2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho)\Big{]}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}\Big{[}z_{i,n}^{+}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})\Big{]}\Big{[}z_{i,n}^{+}+2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho)\Big{]}}\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{+}-1}(z_{i,n}^{+})^{2}}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{z}_{i,n}^{+}(-\bar{z}_{i,n}^{+})},\,\text{if}\,\mathfrak{K}>1,$

and

$\displaystyle\mathcal{G}(i\mathfrak{s}^{\star}\rho)=\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}[\bar{z}_{i}^{+}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})]\prod\limits_{i=1}^{{\boldsymbol{K}}^{-}-1}(z_{i}^{-})^{2}}{\omega_{0}(1-q_{2}^{*}q_{1})(\mathfrak{s}-\mathfrak{s}^{*}\rho)\prod\limits_{i=1}^{{\boldsymbol{K}}^{-}-1}[z_{i}^{-}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})][z_{i}^{-}-2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})]\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{z}_{i}^{+}}$
$\displaystyle\times\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\Big{[}\bar{z}_{i,n}^{+}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})\Big{]}\Big{[}-\bar{z}_{i,-n}^{+}+2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho)\Big{]}}{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}\Big{[}z_{i,n}^{-}+2\omega_{0}(\mathfrak{s}^{\star}\rho-\mathfrak{s})\Big{]}\Big{[}z_{i,n}^{-}+2\omega_{0}(\mathfrak{s}-\mathfrak{s}^{\star}\rho)\Big{]}}\times\frac{\prod\limits_{i=0}^{{\boldsymbol{K}}^{-}-1}(z_{i,n}^{-})^{2}}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{z}_{i,n}^{+}(-\bar{z}_{i,n}^{+})},\,\text{if}\,\mathfrak{K}<1.$

At this point, we state the first assumption on $\mathfrak{s}$.

H15: We require that $\mathfrak{s}$ satisfies inequalities: $\,\sin
2\omega_{0}\mathfrak{s}+\text{sgn}(\mathfrak{K}-1)q^{*}_{2}\sin\pi\mathfrak{s}\neq
0,$ and

$|\mathfrak{s}|<4\pi q,\quad
2\omega_{0}\mathfrak{s}\neq-\bar{z}^{+}_{i,n},\,2\omega_{0}\mathfrak{s}\neq\bar{z}^{+}_{i,-n},\quad
i\in\\{0,1,...,\bar{{\boldsymbol{K}}}^{+}-1\\},\,$

and

$2\omega_{0}\mathfrak{s}\neq\pm z^{-}_{i,n},\,\quad
i\in\\{0,1,...,{\boldsymbol{K}}^{-}-1\\},\,n\in\mathbb{N}\cup\\{0\\},\quad\text{if}\,\mathfrak{K}<1,$

while

$2\omega_{0}\mathfrak{s}\neq\pm
z^{+}_{i,n},\,i\in\\{0,1,...,{\boldsymbol{K}}^{+}-1\\},\,n\in\mathbb{N}\cup\\{0\\},\quad\text{if}\,\mathfrak{K}>1.$

Then, setting for $n\in\mathbb{N}\cup\\{0\\}$

$\bar{{\boldsymbol{K}}}:=\begin{cases}{\boldsymbol{K}}^{+},\quad\text{if}\quad\mathfrak{K}>1,\\\
{\boldsymbol{K}}^{-},\quad\text{if}\quad\mathfrak{K}<1,\end{cases}$
$\bar{Z}^{+}_{i,n}:=\frac{\bar{z}_{i,n}^{+}-2\omega_{0}\mathfrak{s}}{|\mathfrak{s}^{\star}|2\omega_{0}},\quad\bar{Z}^{+}_{i,-n}:=\frac{-\bar{z}_{i,-n}^{+}+2\omega_{0}\mathfrak{s}}{|\mathfrak{s}^{\star}|2\omega_{0}},\quad
i\in\\{0,1,...,\bar{{\boldsymbol{K}}}^{+}-1\\},$
$Z_{i,n}=\begin{cases}\frac{z_{i,n}^{+}-2\omega_{0}\mathfrak{s}}{|\mathfrak{s}^{\star}|2\omega_{0}},\quad\text{if}\quad\mathfrak{K}>1,\\\
\frac{z_{i,n}^{-}-2\omega_{0}\mathfrak{s}}{|\mathfrak{s}^{\star}|2\omega_{0}},\quad\text{if}\quad\mathfrak{K}<1,\end{cases}\quad
Z_{i,-n}=\begin{cases}\frac{z_{i,n}^{+}+2\omega_{0}\mathfrak{s}}{|\mathfrak{s}^{\star}|2\omega_{0}},\quad\text{if}\quad\mathfrak{K}>1,\\\
\frac{z_{i,n}^{-}+2\omega_{0}\mathfrak{s}}{|\mathfrak{s}^{\star}|2\omega_{0}},\quad\text{if}\quad\mathfrak{K}<1,\end{cases}\,i\in\\{0,1,...,\bar{{\boldsymbol{K}}}\\},$

and taking into account H15 and the easy verified equality

$\mathcal{G}(0)=-\sqrt{1+a_{3}^{2}}\frac{\sin(2\omega_{0}\mathfrak{s}-\theta_{1})+q_{2}\sin(\pi\mathfrak{s}-\theta_{2})}{\sin
2\omega_{0}\mathfrak{s}+\text{sgn}(\mathfrak{K}-1)q_{2}^{*}\sin\pi\mathfrak{s}},$

we have

$\displaystyle\mathcal{G}(i\mathfrak{s}^{\star}\rho)$
$\displaystyle=\frac{\mathcal{G}(0)\mathfrak{s}}{(\mathfrak{s}-\mathfrak{s}^{\star}\rho)}\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}[\bar{Z}^{+}_{i,0}+\text{sgn}(\mathfrak{s}^{\star})\rho]}{\prod\limits_{i=1}^{\bar{{\boldsymbol{K}}}-1}[Z_{i,0}+\text{sgn}(\mathfrak{s}^{\star})\rho][Z_{i,-0}-\text{sgn}(\mathfrak{s}^{\star})\rho]}\frac{\prod\limits_{i=1}^{\bar{{\boldsymbol{K}}}-1}Z_{i,0}Z_{i,-0}}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{Z}_{i,0}^{+}}$
$\displaystyle\times\prod\limits_{n=1}^{+\infty}\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}[\bar{Z}^{+}_{i,n}+\text{sgn}(\mathfrak{s}^{\star})\rho][\bar{Z}^{+}_{i,-n}-\text{sgn}(\mathfrak{s}^{\star})\rho]}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}[Z_{i,n}+\text{sgn}(\mathfrak{s}^{\star})\rho][Z_{i,-n}-\text{sgn}(\mathfrak{s}^{\star})\rho]}\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}Z_{i,n}Z_{i,-n}}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{Z}_{i,n}^{+}\bar{Z}_{i,-n}^{+}}.$
(6.10)

In virtue of assumptions H15 and Corollary 5.1, we conclude that the
sequences: $\\{Z_{i,n}\\}_{n=0}^{\infty}$, $\\{Z_{i,-n}\\}_{n=0}^{\infty}$,
$\\{\bar{Z}^{+}_{i,n}\\}_{n=0}^{\infty}$,
$\\{\bar{Z}^{+}_{i,-n}\\}_{n=0}^{\infty}$, meet the requirements of H2 and H3
for each $i$, besides the terms of these sequences are positive for $n\geq 1$
and don’t vanish if $n=0$.

Now, coming to equation (6.8) and keeping in mind factorization (6), we employ
Theorems 2.1 and 2.2 to build the explicit solutions homogeneous and
inhomogeneous equation (6.8).

At this point, we first find a solution of homogeneous (6.11)
($\mathcal{F}\equiv 0$). To this end, putting

$\mathfrak{d}_{0}(\mathfrak{K}):=\frac{\mathcal{G}(0)\mathfrak{s}\prod\limits_{i=1}^{\bar{{\boldsymbol{K}}}-1}Z_{i,0}Z_{i,-0}}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{Z}_{i,0}^{+}},\,\,\mathbb{L}_{0}(\rho,\mathfrak{K},\mathfrak{s}^{\star})=\begin{cases}\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\Gamma(\bar{Z}^{+}_{i,0}+\rho)\prod\limits_{i=1}^{\bar{K}-1}\frac{\Gamma(1+Z_{i,-0}-\rho)}{\Gamma(\rho+Z_{i,0})},\quad\text{if}\quad\mathfrak{s}^{\star}>0,\\\
\frac{1}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\Gamma(1+\bar{Z}^{+}_{i,0}-\rho)}\prod\limits_{i=1}^{\bar{K}-1}\frac{\Gamma(1+Z_{i,0}-\rho)}{\Gamma(\rho+Z_{i,-0})},\quad\,\text{if}\quad\mathfrak{s}^{\star}<0,\end{cases}$
$\mathcal{R}(n,\mathfrak{K},\mathfrak{s}^{\star})=\begin{cases}\sum\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}[\bar{Z}_{i,-n}^{+}(\ln\bar{Z}_{i,-n}^{+}-1)-\bar{Z}_{i,n}^{+}(\ln\bar{Z}_{i,n}^{+}-1)]\\\
-\sum\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}[Z_{i,-n}(\ln
Z_{i,-n}-1)-Z_{i,n}(\ln Z_{i,n}-1)],\quad\text{ if
}\quad\mathfrak{s}^{\star}>0,\\\
\sum\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}[\bar{Z}_{i,-n}^{+}(\ln\bar{Z}_{i,-n}^{+}-1)-\bar{Z}_{i,n}^{+}(\ln\bar{Z}_{i,n}^{+}-1)]\\\
+\sum\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}[Z_{i,-n}(\ln
Z_{i,-n}-1)-Z_{i,n}(\ln Z_{i,n}-1)],\quad\text{ if
}\quad\mathfrak{s}^{\star}<0,\end{cases}$

and

$\mathbb{L}_{1}(\rho,\mathfrak{K},\mathfrak{s}^{\star})=\begin{cases}\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\frac{\Gamma(\bar{Z}^{+}_{i,n}+\rho)}{\Gamma(\bar{Z}^{+}_{i,-n}+1-\rho)}\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}\frac{\Gamma(Z_{i,-n}-\rho+1)}{\Gamma(Z_{i,n}+\rho)}\bigg{(}\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}Z_{i,n}Z_{i,-n}}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{Z}^{+}_{i,n}\bar{Z}^{+}_{i,-n}}\bigg{)}^{\rho-\frac{1}{2}}e^{\mathcal{R}(\rho,\mathfrak{K},\,\mathfrak{s}^{\star})}\quad\text{if}\quad\mathfrak{s}^{\star}>0,\\\
\prod\limits_{n=1}^{+\infty}\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\frac{\Gamma(\bar{Z}^{+}_{i,-n}+\rho)}{\Gamma(\bar{Z}^{+}_{i,n}+1-\rho)}\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}\frac{\Gamma(Z_{i,n}-\rho+1)}{\Gamma(Z_{i,-n}+\rho)}\bigg{(}\frac{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}-1}Z_{i,n}Z_{i,-n}}{\prod\limits_{i=0}^{\bar{{\boldsymbol{K}}}^{+}-1}\bar{Z}^{+}_{i,n}\bar{Z}^{+}_{i,-n}}\bigg{)}^{\rho-\frac{1}{2}}e^{\mathcal{R}(\rho,\mathfrak{K},\,\mathfrak{s}^{\star})}\quad\text{if}\quad\mathfrak{s}^{\star}<0,\end{cases}$

and exploiting Theorem 2.1, we end up with the explicit form of the solution
to homogeneous equation (6.8):

$V_{h}(\rho,\sigma;{\mathbb{P}})=(\mathfrak{d}_{0}(\mathfrak{K}))^{\rho-1/2}{\mathbb{P}}(\rho)(a_{1}\sigma+a_{2}\sigma^{\nu})^{\frac{1}{2}-\rho}\mathbb{L}_{0}(\rho,\mathfrak{K},\mathfrak{s}^{\star})\mathbb{L}_{0}(\rho,\mathfrak{K},\mathfrak{s}^{\star})$
(6.11)

where, for $m\in\mathbb{N}\cup\\{0\\}$ and $n\in\mathbb{N},$ $Re\,\rho$
satisfies the following inequalities (similar to (2.2) and (2.3)) :

$Re\rho\neq-m-\bar{Z}_{i,n}^{+},\,i\in\\{0,...,\bar{{\boldsymbol{K}}}^{+}-1\\},\,Re\rho\neq
m+1+Z_{j,1-n},\,j\in\\{0,...,\bar{{\boldsymbol{K}}}-1\\},\,Re\rho\neq
1+m+Z_{0,-n}$ (6.12)

if $\mathfrak{s}^{\star}>0$, while

$Re\rho\neq-m-\bar{Z}_{i,-n}^{+},\,i\in\\{0,...,\bar{{\boldsymbol{K}}}^{+}-1\\},\,Re\rho\neq
m+1+Z_{j,n-1},\,j\in\\{0,...,\bar{{\boldsymbol{K}}}-1\\},\,Re\rho\neq
1+m+Z_{0,n}$ (6.13)

if $\mathfrak{s}^{\star}<0$.

Besides, the following statement is simple consequence of Theorem 2.1.

######  Proposition 6.1.

Let $Re\,\rho$ meet requirement (6.12) in the case of $\mathfrak{s}^{\star}>0$
and (6.13) if $\mathfrak{s}^{\star}<0$. Then the function
$\frac{V_{h}(\rho,\sigma;{\mathbb{P}})}{{\mathbb{P}}(\rho)}$ has no poles and
the infinite product $\mathbb{L}_{1}(\rho,\mathfrak{K},\mathfrak{s}^{\star})$
converges for each complex $\rho$. Besides, the relations hold:

$\displaystyle(\mathfrak{d}_{0}(\mathfrak{K}))^{\rho-1/2}\mathbb{L}_{0}(\rho,\mathfrak{K},\mathfrak{s}^{\star})\mathbb{L}_{0}(\rho,\mathfrak{K},\mathfrak{s}^{\star})=[(\mathfrak{s}-\mathfrak{s}^{\star}\rho)\mathcal{G}(i\mathfrak{s}^{\star}\rho)]^{\rho-1/2}\exp\\{C_{1}\ln\rho+C_{2}\rho+O(1)\\},$
$\displaystyle|V_{h}(\rho,\sigma;{\mathbb{P}}){\mathbb{P}}^{-1}(\rho)|\leq
C|\rho|^{C_{1}+Re\rho-1/2}\exp\\{\arg(a_{1}\sigma+a_{2}\sigma^{\nu})Im\rho+[-\frac{\pi}{2}\text{sgn}\,(Re\rho)+\theta_{2}(\text{sgn}\,\mathfrak{s}^{\star})]|Im\rho|\\},$

as $|Im\rho|\to+\infty$ and $Re\rho$ remains bounded.

Since we look for a bounded solution at the infinity, Remark 2.1 and Theorem
2.2 tell us that we are left to construct the particular solution in form
(2.10). To this end, we first put the function
${\mathbb{P}}(\rho):={\mathbb{P}}_{1}(\rho)$ where

${\mathbb{P}}_{1}(\rho)=\begin{cases}\prod\limits_{i=1}^{5}\sin\pi(1+Z_{i,0}-\rho)\prod\limits_{i=0}^{2}\sin^{-1}\pi(1+\bar{Z}_{i,0}^{+}-\rho),\quad\text{if}\,\mathfrak{s}^{\star}<0,\\\
\prod\limits_{i=0}^{4}\sin\pi(\bar{Z}_{i,0}^{+}+\rho)\prod\limits_{i=1}^{3}\sin^{-1}\pi(Z_{i,0}+\rho),\qquad\qquad\,\text{
if }\mathfrak{s}^{\star}>0.\end{cases}$ (6.14)

It is apparent that the introduced function ${\mathbb{P}}_{1}$ obeys the
properties:

${\mathbb{P}}_{1}(\rho)\approx
Ce^{2\pi|Im\rho|},\quad\text{as}\quad|Im\rho|\to+\infty,$ (6.15)

besides the functions:
$\frac{{\mathbb{P}}_{1}(\rho)\prod\limits_{i=1}^{5}\Gamma(1+Z_{i,0}-\rho)}{\prod\limits_{i=0}^{2}\Gamma(1+\bar{Z}_{i,0}^{+}-\rho)}\,$
if $\mathfrak{s}^{\star}<0\,$ and
$\frac{{\mathbb{P}}_{1}(\rho)\prod\limits_{i=0}^{4}\Gamma(\bar{Z}_{i,0}^{+}+\rho)}{\prod\limits_{i=1}^{3}\Gamma(Z_{i,0}+\rho)}\,$
if $\mathfrak{s}^{\star}>0$ do not have any poles if for
$l_{1},l_{2}\in\mathbb{N}$, the inequalities hold

$Re\rho\neq l_{2}-Z_{i,0},\,i\in\\{1,2,3\\},\text{ if
}\mathfrak{s}^{\star}>0,\,\text{and }Re\rho\neq
1+\bar{Z}^{+}_{j,0}-l_{1},\,j\in\\{0,1,2\\},\text{ if
}\mathfrak{s}^{\star}<0,$ (6.16)

and do not have any nulls if

$Re\rho\neq l_{2}-\bar{Z}^{+}_{i,0},\,i\in\\{0,1,2,3,4,\\}\text{ if
}\mathfrak{s}^{\star}>0,\,\text{and }Re\rho\neq
1+Z_{j,0}-l_{1},\,j\in\\{1,2,3,4,5\\}\text{ if }\mathfrak{s}^{\star}<0.$
(6.17)

Collecting Proposition 6.1 with (6.12)-(6.17) and setting

$Z^{*}=\begin{cases}1+Z_{6,-0},\quad\text{if}\quad\bar{{\boldsymbol{K}}}>6,\\\
4\pi q,\quad\qquad\text{ if}\quad\bar{{\boldsymbol{K}}}\leq 6,\end{cases}$

we conclude.

######  Proposition 6.2.

The function $V_{h}(\rho,\sigma):=V_{h}(\rho,\sigma;{\mathbb{P}}_{1})$ given
with (6.11) and (6.14) has the following properties

i:

$V_{h}(\rho,\sigma)$ does not have any poles if in the case of
$\mathfrak{s}^{\star}<0$ the inequalities hold

$-\bar{Z}_{0,-1}^{+}<Re\rho<Z^{*}\,\text{and }Re\rho\neq
1-\bar{m}(i)+\bar{Z}^{+}_{i,0},\quad i\in\\{0,1,2\\},$ (6.18)

where the integer $\bar{m}(i)$ satisfies inequalities

$\max\\{1;\bar{Z}_{i,0}^{+}-Z^{*}+\\}<\bar{m}(i)<1+\bar{Z}_{i,0}^{+}+\bar{Z}_{0,-1}^{+},$

while in the case of $\mathfrak{s}^{\star}>0$ there hold

$-\bar{Z}_{5,0}^{+}<Re\rho<1+Z_{1,-0},\,Re\rho\neq\bar{m}(i)-Z_{i,0},\quad
i\in\\{1,2,3\\},$ (6.19)

where the integer $\bar{m}(i)$ satisfies relations

$\max\\{1;Z_{i,0}-\bar{Z}^{+}_{5,0}\\}<\bar{m}(i)<1+Z_{i,0}+Z_{1,-0};$

ii:

$V_{h}(\rho,\sigma)$ does not vanish if in the case of
$\mathfrak{s}^{\star}<0$ the inequalities hold:

$-{Z}_{1,-0}<Re\rho<1+\bar{Z}^{+}_{3,0},\text{ and }\,Re\rho\neq
1-\underline{m}(i)+Z_{i,0},\quad i\in\\{1,2,3,4,5\\},$ (6.20)

with the integer $\underline{m}(i)$ satisfying bounds:
$\max\\{1;Z_{i,0}-\bar{Z}^{+}_{3,0}\\}<\underline{m}(i)<1+Z_{i,0}+Z_{1,-0},$
while in the case of $\mathfrak{s}^{\star}>0$ there are inequalities

$-Z_{4,0}<Re\rho<1+\bar{Z}^{+}_{0,-1},\text{ and
}\,Re\rho\neq\underline{m}(i)-\bar{Z}^{+}_{i,0},\quad i\in\\{0,1,2,3,4\\},$
(6.21)

with the integer $\underline{m}(i)$ satisfying bounds:
$\max\\{1;\bar{Z}^{+}_{i,0}-Z_{4,0}\\}<\underline{m}(i)<1+\bar{Z}^{+}_{i,0}+\bar{Z}^{+}_{0,-1}.$

iii:

If $Re\rho$ meets requirements (6.20) and (6.21), then the estimate holds

$|V_{h}(\rho,\sigma)|^{-1}\leq
C\exp\\{-\Big{[}\frac{\pi}{2}-\varepsilon+\theta_{2}(sgn\mathfrak{s}^{\star})\Big{]}|Im\rho|\\},\quad\text{as}\quad|Im\rho|\to+\infty,$

where $\varepsilon$ is a sufficiently small positive number,
$0<\varepsilon<\theta/10$.

Taking into account Propositions 6.1 and 6.2 and Theorem 2.2, and choosing
$\mathcal{K}_{1}(\xi)=\frac{\sin^{2}\pi d}{\sin^{2}\pi(\xi+d)},$ $0<d-d_{0}<1$
(see Table LABEL:tab:table5 in Example 5.2), we construct the particular
solution of equation (6.8).

######  Proposition 6.3.

Let assumptions H14 and H15 hold. Moreover, we assume that

$-Z_{1,-0}<Re\rho<\bar{Z}_{3,0}^{+},\text{ and }\,Re\rho\neq
d_{0}+Z_{i,-0}-\underline{m}(i),\quad i\in\\{1,2,3,4,5\\},$

if $\mathfrak{s}^{\star}<0$, and in the case of $\mathfrak{s}^{\star}>0$,
there are

$-Z_{4,0}<Re\rho<\bar{Z}_{0,-1}^{+},\text{ and }\,Re\rho\neq
d_{0}-Z_{i,0}-1+\underline{m}(i),\quad i\in\\{0,1,2,3,4\\},$

with $\underline{m}(i)$ meets the requirements of (ii) in Proposition 6.2.
Then the function

$V(\rho,\sigma)=\frac{1}{2i}\int\limits_{\ell_{d_{0}}}\frac{V_{h}(\rho,\sigma;{\mathbb{P}}_{1})f^{*}(i\mathfrak{s}^{\star}\rho+i\mathfrak{s}^{\star}\xi+i(1-\mathfrak{s}),\sigma)}{V_{h}(\rho+1+\xi,\sigma;{\mathbb{P}}_{1})(a_{1}\sigma+a_{2}\sigma^{\nu})}[\cot\pi\xi+i]\frac{\sin^{2}\pi
d}{\sin^{2}\pi(\xi+d)}d\xi$

solves equation (6.8). Besides, the function
$\frac{V_{h}(\rho,\sigma;{\mathbb{P}}_{1})}{V_{h}(\rho+1+\xi,\sigma;{\mathbb{P}}_{1})}$
is analytic in $\rho$ if $\xi=-d_{0}+iy,$ $y\in\mathbb{R}$, $d_{0}\in[0,1],$
and $Re\rho$ meets the requirements

$-Z_{1,-0}<Re\rho<\bar{Z}^{+}_{3,0},\,\,Re\rho\neq
d_{0}+Z_{i,-0}-\underline{m}(i),\,\,i\in\\{1,2,3,4,5\\},\,\,Re\rho\neq
1-\bar{m}(i)+\bar{Z}_{j,0}^{+},\,\,dj\in\\{0,1,2\\},$

in the case of $\mathfrak{s}^{\star}<0,$ and

$-Z_{4,0}<Re\rho<1+Z_{1,-0},\,\,Re\rho\neq
d_{0}-Z_{i,0}-1+\underline{m}(i),\,\,i\in\\{0,1,2,3,4\\},\,\,Re\rho\neq\bar{m}(i)-Z_{j,0},\,\,j\in\\{1,2,3\\},$

if $\mathfrak{s}^{\star}>0.$ Integer $\underline{m}(i)$ and $\bar{m}(i)$ meet
requirement of Proposition 6.2.

Third Stage. Denoting

$\mathcal{N}_{2}(\mathfrak{x})=[\cot\mathfrak{x}(\omega_{0}-\pi/2)+a_{4}][\cot\mathfrak{x}(\omega_{0}+\pi/2)+\mathfrak{K}a_{4}]^{-1},$

and collecting Proposition 6.3 with relations (6.5) and (6.6), (6.7), we end
up with the solution

$\displaystyle U_{1}^{*}(\lambda,x_{2},\sigma)$
$\displaystyle=\frac{V_{h}(-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})}{2i}\frac{\mathfrak{K}\mathcal{N}_{2}(i\lambda+\mathfrak{s})\sin(i\lambda+\mathfrak{s})(x_{2}+\pi/2)}{[\mathfrak{K}\mathcal{N}_{2}(i\lambda+\mathfrak{s})+1]\sin(i\lambda+\mathfrak{s})(\omega_{0}+\pi/2)}$
$\displaystyle\times\int\limits_{\ell_{d_{0}}}\frac{f^{*}(\lambda+i\mathfrak{s}^{\star}\xi+i(1-\mathfrak{s}),\sigma)[\cot\pi\xi+i]}{(a_{1}\sigma+a_{2}\sigma^{\nu})V_{h}(1+\xi-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})}\frac{\sin^{2}\pi
d}{\sin^{2}\pi(\xi+d)}d\xi,$ $\displaystyle U_{2}^{*}(\lambda,x_{2},\sigma)$
$\displaystyle=\frac{V_{h}(-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})}{2i}\frac{\sin(i\lambda+\mathfrak{s})(x_{2}-\pi/2)}{[\mathfrak{K}\mathcal{N}_{2}(i\lambda+\mathfrak{s})+1]\sin(i\lambda+\mathfrak{s})(\omega_{0}-\pi/2)}$
(6.22)
$\displaystyle\times\int\limits_{\ell_{d_{0}}}\frac{f^{*}(\lambda+i\mathfrak{s}^{\star}\xi+i(1-\mathfrak{s}),\sigma)[\cot\pi\xi+i]}{(a_{1}\sigma+a_{2}\sigma^{\nu})V_{h}(1+\xi-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})}\frac{\sin^{2}\pi
d}{\sin^{2}\pi(\xi+d)}d\xi.$

Finally, we are left to compute inverse Laplace and Fourier transformations to
obtain the integral representation of the solution (1.5)-(1.9). To this end,
we appeal to (5.1.33) in [14] and deduce

$\mathcal{L}(t,y):=\frac{1}{2\pi
i}\int\limits_{-i\infty}^{+i\infty}\frac{e^{\sigma
t}d\sigma}{(a_{1}\sigma+a_{2}\sigma^{\nu})^{d_{0}-iy}}=\begin{cases}\frac{t^{-1+\frac{a_{1}+a_{2}\nu}{a_{1}+a_{2}}[d_{0}-iy]}}{\Gamma\big{(}\frac{a_{1}+a_{2}\nu}{a_{1}+a_{2}}[d_{0}-iy]\big{)}(a_{2}+a_{1})^{d_{0}-iy}},\quad\quad\text{if}\,\text{
either }a_{1}=0\text{ or }a_{2}=0,\\\ \,\\\
a_{1}^{-d_{0}+iy}t^{-1+d_{0}-iy}E_{1-\nu,d_{0}-iy}^{d_{0}-iy}(-\frac{a_{2}}{a_{1}}t^{1-\nu}),\quad\text{if}\quad
a_{1},a_{2}>0,\end{cases}$

where $E_{\alpha,\beta}^{\gamma}(\cdot)$ is the three-parametric Mittag-
Leffler function (see, e.g. (5.1.4) in [14]).

Then, collecting these relations with (6.22), we arrive at the explicit
solution to (1.5)-(1.9).

######  Proposition 6.4.

Let assumptions of Propositions 6.1-6.3 and restriction (6.1) hold, then the
following functions solves problem (1.5)-(1.9):

$\displaystyle U_{1}(x_{1},x_{2},t)$
$\displaystyle=\int\limits_{0}^{t}d\tau\int\limits_{-\infty}^{+\infty}d\varrho
f(x_{1}-\varrho,t-\tau)e^{[-\mathfrak{s}^{\star}d_{0}+1-\mathfrak{s}](x_{1}-\varrho)}\int\limits_{-\infty}^{+\infty}dy\frac{\mathcal{L}(t,y)e^{iy\mathfrak{s}^{\star}(x_{1}-\varrho)}\sin^{2}\pi
d}{\sin\pi(d_{0}-iy)\sin^{2}\pi(d-d_{0}+iy)}$
$\displaystyle\times\int\limits_{-\infty}^{+\infty}\frac{e^{i\lambda\varrho}V_{h}(-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})\mathfrak{K}\mathcal{N}_{2}(i\lambda+\mathfrak{s})\sin(i\lambda+\mathfrak{s})(x_{2}+\pi/2)}{V_{h}(1-d_{0}+iy-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})[1+\mathfrak{K}\mathcal{N}_{2}(i\lambda+\mathfrak{s})]\sin(i\lambda+\mathfrak{s})(\omega_{0}+\pi/2)}d\lambda;$
$\displaystyle U_{2}(x_{1},x_{2},t)$
$\displaystyle=\int\limits_{0}^{t}d\tau\int\limits_{-\infty}^{+\infty}d\varrho
f(x_{1}-\varrho,t-\tau)e^{[-\mathfrak{s}^{\star}d_{0}+1-\mathfrak{s}](x_{1}-\varrho)}\int\limits_{-\infty}^{+\infty}dy\frac{\mathcal{L}(t,y)e^{iy\mathfrak{s}^{\star}(x_{1}-\varrho)}\sin^{2}\pi
d}{\sin\pi(d_{0}-iy)\sin^{2}\pi(d-d_{0}+iy)}$
$\displaystyle\times\int\limits_{-\infty}^{+\infty}\frac{e^{i\lambda\varrho}V_{h}(-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})\sin(i\lambda+\mathfrak{s})(x_{2}-\pi/2)}{V_{h}(1-d_{0}+iy-i\lambda/\mathfrak{s}^{\star},\sigma;{\mathbb{P}}_{1})[1+\mathfrak{K}\mathcal{N}_{2}(i\lambda+\mathfrak{s})]\sin(i\lambda+\mathfrak{s})(\omega_{0}-\pi/2)}d\lambda,$

where $Im\lambda$ meets requirements:

$-Z_{1,-0}<\frac{Im\lambda}{-\mathfrak{s}^{\star}}<\bar{Z}^{+}_{3,0},\,\,\frac{Im\lambda}{\mathfrak{s}^{\star}}\neq
d_{0}+Z_{i,-0}-\underline{m}(i),\,i\in\\{1,2,3,4,5\\},\,\,\frac{Im\lambda}{\mathfrak{s}^{\star}}\neq
1-\bar{m}(j)+\bar{Z}_{j,0}^{+},\,j\in\\{0,1,2\\},$

in the case of $\mathfrak{s}^{\star}<0$, while if $\mathfrak{s}^{\star}>0$
then the inequalities hold

$-Z_{4,0}<\frac{Im\lambda}{\mathfrak{s}^{\star}}<1+Z_{1,-0},\,\,\frac{Im\lambda}{\mathfrak{s}^{\star}}\neq
d_{0}-1-Z_{i,0}+\underline{m}(i),\,i\in\\{0,1,2,3,4\\},\,\,\frac{Im\lambda}{\mathfrak{s}^{\star}}\neq\bar{m}(j)-Z_{j,0},\,j\in\\{1,2,3\\}$

with integer $\underline{m}(i)$ and $\bar{m}(i)$ satisfying relations in
(i),(ii) of Proposition 6.2.

######  Remark 6.2.

It is worth noting that relations on $Im\lambda$ in Proposition 6.4 provide
additional assumptions on the weight $\mathfrak{s}$. Indeed, putting
$Im\lambda=0$ in the relations above, we arrive at the conditions:

$\bullet$ if $\mathfrak{s}^{\star}<0$, then
$\mathfrak{s}\neq\frac{\bar{z}_{i}^{+}}{2\omega_{0}}-|\mathfrak{s}^{\star}|[\bar{m}(i)-1],\,i\in\\{0,1,2\\},$
and for $j\in\\{0,1,...,5\\}$

$-\frac{z_{1}^{+}}{2\omega_{0}}<\mathfrak{s}<\frac{\bar{z}_{3}^{+}}{2\omega_{0}},\quad\mathfrak{s}\neq|\mathfrak{s}^{\star}|[\underline{m}(j)-d_{0}]-\frac{z_{j}^{+}}{2\omega_{0}},\,\,\text{
if }\,\mathfrak{K}>1,$
$-\frac{z_{1}^{-}}{2\omega_{0}}<\mathfrak{s}<\frac{\bar{z}_{3}^{+}}{2\omega_{0}},\quad\mathfrak{s}\neq|\mathfrak{s}^{\star}|[\underline{m}(j)-d_{0}]-\frac{z_{j}^{-}}{2\omega_{0}},\,\,\text{
if }\,\mathfrak{K}<1;$

$\bullet$ if $\mathfrak{s}^{\star}>0$, then

$-\frac{z_{1}^{+}}{2\omega_{0}}-1<\mathfrak{s}<\frac{z_{4}^{+}}{2\omega_{0}},\mathfrak{s}\neq\frac{z_{i}^{+}}{2\omega_{0}}+|\mathfrak{s}^{\star}|[1-\underline{m}(i)-d_{0}],\,i\in\\{0,...,4\\},\mathfrak{s}\neq-|\mathfrak{s}^{\star}|\bar{m}(j)+\frac{z_{j}^{+}}{2\omega_{0}},j\in\\{1,2,3\\},\text{if
}\mathfrak{K}>1,$

while if $\mathfrak{K}<1$ then

$-\frac{z_{1}^{-}}{2\omega_{0}}-1<\mathfrak{s}<\frac{z_{4}^{-}}{2\omega_{0}},\mathfrak{s}\neq\frac{z_{i}^{-}}{2\omega_{0}}+|\mathfrak{s}^{\star}|[1-\underline{m}(i)-d_{0}],\,i\in\\{0,...,4\\},\mathfrak{s}\neq-|\mathfrak{s}^{\star}|\bar{m}(j)+\frac{z_{j}^{-}}{2\omega_{0}},j\in\\{1,2,3\\}.$

It is apparent that the selected weight $\mathfrak{s}$ in accordance with
these assumptions will satisfy H15.

We conclude this section with some comments related to other boundary problems
with dynamic boundary conditions in plane corners.

######  Remark 6.3.

Actually, with nonessential modifications in the arguments in Section 6, the
very same results hold for transmission problems (1.5)-(1.8) supplemented with
Neumann boundary conditions on $\partial G_{i}\backslash g,$ $i=1,2.$ Besides,
the proposed approach in Section 6 can be incorporated to find explicit
solutions of the non-classical Dirichlet or Neumann boundary value problem
with fractional dynamic boundary conditions in plane corners.

## Appendix: Proof of Proposition 3.1

We start with validation of the first estimate in Proposition 3.1. To this
end, it is enough to evaluate the first term in the left-hand side of the
inequality (i). The second series will be examined with the same arguments.
The simple technical calculations and properties of the sequence of
$\mathfrak{b}(n)$ provide the relations

$\displaystyle\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{z}{\mathfrak{b}(n)(\mathfrak{b}(n)+z)}=\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{z_{1}(\mathfrak{b}(n)+z_{1})}{\mathfrak{b}(n)[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}]}+i\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{z_{2}}{[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}]}$
$\displaystyle+\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{z_{2}^{2}}{\mathfrak{b}(n)[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}]}=C+O(1/|z_{2}|)+\sum_{n=\mathfrak{N}+1}^{+\infty}\frac{z_{2}^{2}}{\mathfrak{b}(n)[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}]}.$
(A.1)

To handle the last term in the right-hand side of this equality, we take
advantage of the easy verified inequalities

$\Big{|}\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{z_{2}^{2}}{\mathfrak{b}(n)[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}]}\Big{|}\leq\int_{\mathfrak{N}\,+1}^{+\infty}\frac{Cz_{2}^{2}dx}{\mathfrak{b}(x)[z_{2}^{2}+\mathfrak{b}^{2}(x)]}\leq\frac{z_{2}^{2}}{\delta_{0}}\int_{\mathfrak{b}(\mathfrak{N}\,+1)}^{+\infty}\frac{Cdy}{y[z_{2}^{2}+y^{2}]}$

with $\delta_{0}>0$. Here we used that the sequence
$\\{\mathfrak{b}(n)\\}_{n=1}^{+\infty}$ is the strictly increasing, i.e.

$\mathfrak{b}^{\prime}(x)>\delta_{0}>0,\quad\forall x\geq\mathfrak{N}+1.$
(A.2)

Performing the change of variable $(z^{2}_{2}y^{-2})=\mathfrak{p}$, we obtain

$\Big{|}\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{z_{2}^{2}}{\mathfrak{b}(n)[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}]}\Big{|}\leq\frac{C}{\delta_{0}}\int^{z_{2}^{2}\mathfrak{b}^{-2}(\mathfrak{N}\,+1)}_{0}\frac{d\mathfrak{p}}{1+\mathfrak{p}^{2}}=C\ln[1+z_{2}^{2}\mathfrak{b}^{-2}(\mathfrak{N}\,+1)].$

Collecting this inequality with (Appendix: Proof of Proposition 3.1), we
arrive at the first estimate in Proposition 3.1.

$\bullet$ Concerning the second estimate in Proposition 3.1, we can conclude
that second and third terms in the left-hand side of this inequality are
evaluated exactly like the first one. Hence, we focus here only on the proof
of the equality

$\sum_{n=\mathfrak{N}\,+1}^{+\infty}\Big{|}\frac{z}{(\mathfrak{b}(n)+z)(\mathfrak{b}(n)+C^{\star}+z)}\Big{|}=C+O(1/|z|),$

if $|z_{2}|\to+\infty$ and $z_{1}$ meets the requirements of Proposition 3.1.

Performing the simple calculations, we reach to the representation

$\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{z}{(\mathfrak{b}(n)+z)(\mathfrak{b}(n)+C^{\star}+z)}=z\sum_{j=1}^{4}\mathfrak{B}_{j}(z),$
(A.3)

where we put

$\displaystyle\mathfrak{B}_{1}(z)$
$\displaystyle=\sum_{n=\mathfrak{N}\,+1}\frac{C^{\star}(\mathfrak{b}(n)+z_{1})}{[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}][(\mathfrak{b}(n)+C^{\star}+z_{1})^{2}+z_{2}^{2}]};$
$\displaystyle\mathfrak{B}_{2}(z)$
$\displaystyle=\sum_{n=\mathfrak{N}\,+1}\frac{(\mathfrak{b}(n)+z_{1})^{2}}{[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}][(\mathfrak{b}(n)+C^{\star}+z_{1})^{2}+z_{2}^{2}]};$
$\displaystyle\mathfrak{B}_{3}(z)$
$\displaystyle=-\sum_{n=\mathfrak{N}\,+1}\frac{z_{2}^{2}}{[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}][(\mathfrak{b}(n)+C^{\star}+z_{1})^{2}+z_{2}^{2}]};$
$\displaystyle\mathfrak{B}_{4}(z)$
$\displaystyle=-i\sum_{n=\mathfrak{N}\,+1}\frac{z_{2}(2\mathfrak{b}(n)+2z_{1}+C^{\star})}{[(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}][(\mathfrak{b}(n)+C^{\star}+z_{1})^{2}+z_{2}^{2}]}.$

Next, we exploit the inequality (A.2) to achieve the relations

$\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{1}{\mathfrak{b}^{2}(n)+z_{2}^{2}}\leq\int_{\mathfrak{N}\,+1}^{+\infty}\frac{dx}{\mathfrak{b}^{2}(x)+z_{2}^{2}}\leq\frac{1}{\delta_{0}}\int_{\mathfrak{b}(\mathfrak{N}\,+1)}^{+\infty}\frac{dy}{y^{2}+z_{2}^{2}}=\frac{C}{|z_{2}|},$

which in turn arrive at the bound

$|z_{2}||\mathfrak{B}_{1}(z)|+|\mathfrak{B}_{2}(z)|+|\mathfrak{B}_{3}(z)|+|\mathfrak{B}_{4}(z)|\leq\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{C}{(\mathfrak{b}(n)+C^{\star}+z_{1})^{2}+z_{2}^{2}}\leq\frac{C}{z_{2}^{2}}.$

Collecting these estimates with representation (A.3), we end up with the
desired bound.

$\bullet$ As for the third estimate in Proposition 3.1, to verify this
statement is enough to obtain the inequality

$\sum_{n=\mathfrak{N}+1}\Big{|}\mathfrak{b}(n)\frac{z^{3}}{(\mathfrak{b}(n)+z)^{3}}\sum_{j=1}^{+\infty}\frac{1}{(j+3)}\frac{z^{j}}{(\mathfrak{b}(n)+z)^{j}}\Big{|}\leq
C|z^{2}|[\ln|z|+1]+O(1),\quad\text{as }|z_{2}|\to+\infty$ (A.4)

and $z_{1}$ satisfying the requirements of Proposition 3.1.

It is apparent that assumptions on the sequence $\mathfrak{b}(n)$ and $z_{1}$
provide the bound

$\Big{|}\frac{z}{\mathfrak{b}(n)+z}\Big{|}<1\quad\text{for each
}n\geq\mathfrak{N}+1\quad\text{and }|z_{2}|\to+\infty.$

In light of the last inequality, we have

$\displaystyle\frac{1}{|\mathfrak{b}(n)+z|}\sum_{j=1}^{+\infty}\frac{1}{(j+3)}\Big{|}\frac{z}{(\mathfrak{b}(n)+z)}\Big{|}^{j}\leq\frac{1}{|\mathfrak{b}(n)+z|}\sum_{j=1}^{+\infty}\Big{|}\frac{z}{(\mathfrak{b}(n)+z)}\Big{|}^{j}=\frac{1}{|\mathfrak{b}(n)+z|-|z|}$
$\displaystyle=\frac{\sqrt{(\mathfrak{b}(n)+z_{1})^{2}+z_{2}^{2}}+\sqrt{z_{1}^{2}+z_{2}^{2}}}{\mathfrak{b}(n)(\mathfrak{b}(n)+2z_{1})}\leq
C\sum_{k=1}^{3}\mathfrak{A}_{k}(z),$

where we set

$\mathfrak{A}_{1}(z)=\frac{|z_{1}|}{\mathfrak{b}(n)|\mathfrak{b}(n)+2z_{1}|},\,\mathfrak{A}_{2}(z)=\frac{|z_{2}|}{\mathfrak{b}(n)|\mathfrak{b}(n)+2z_{1}|},\,\mathfrak{A}_{3}(z)=\frac{|\mathfrak{b}(n)+z_{1}|}{\mathfrak{b}(n)|\mathfrak{b}(n)+2z_{1}|}.$

Summarizing, we are left to produce the suitable bound of
$\sum_{n=\mathfrak{N}+1}^{+\infty}|z|\mathfrak{b}(n)\Big{|}\frac{z}{\mathfrak{b}(n)+z}\Big{|}^{2}\sum_{k=1}^{3}\mathfrak{A}_{k}(z).$
To this end, we appeal to estimates in points (i) and (ii) in Proposition 3.1
and arrive at the relations

$\displaystyle\sum_{n=\mathfrak{N}\,+1}^{+\infty}\mathfrak{b}(n)\Big{|}\frac{z}{\mathfrak{b}(n)+z}\Big{|}^{2}[|z|\mathfrak{A}_{1}(z)+\mathfrak{A}_{2}(z)]\leq
C|z|\ln|z|+O(1),$
$\displaystyle\sum_{n=\mathfrak{N}\,+1}^{+\infty}|z|\mathfrak{b}(n)\Big{|}\frac{z}{\mathfrak{b}(n)+z}\Big{|}^{2}\mathfrak{A}_{3}(z)\leq
C|z|[|z|+1].$

in the case of $z_{2}\to\pm\infty$. Collecting these inequalities, we deduce
bound (A.4) and, accordingly, the desired estimate in (iii).

$\bullet$ At this point, we will examine the first equality in (iv). Note that
the second one is verified with the same arguments. First, we use the easily
verified relations:

$\displaystyle\sum_{j=1}^{+\infty}\frac{1}{(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j}$
$\displaystyle=\sum_{j=1}^{+\infty}\frac{-3}{j(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j}+\sum_{j=1}^{+\infty}\frac{1}{j}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j},$
$\displaystyle\sum_{j=1}^{+\infty}\frac{1}{(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{j}$
$\displaystyle=\sum_{j=1}^{+\infty}\frac{-3}{j(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{j}+\sum_{j=1}^{+\infty}\frac{1}{j}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{j}.$

In particular, taking into account of the assumptions on $\mathfrak{b}(n)$ and
$z_{1}$, we deduce the equalities:

$\displaystyle\sum_{j=1}^{+\infty}\frac{1}{(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j}$
$\displaystyle=\sum_{j=1}^{+\infty}\frac{-3}{j(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j}-\ln\Big{(}1-\frac{z}{\mathfrak{b}(n)+z}\Big{)},$
$\displaystyle\sum_{j=1}^{+\infty}\frac{1}{(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{j}$
$\displaystyle=\sum_{j=1}^{+\infty}\frac{1}{j(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{j}-\ln\Big{(}1-\frac{z}{\mathfrak{b}(n)+z+C^{\star}}\Big{)}.$

Therefore, we end up with the equality

$\displaystyle\sum_{n=\mathfrak{N}\,+1}^{\infty}\Big{[}\mathfrak{b}(n)\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{3}\sum_{j=1}^{+\infty}\frac{1}{j+3}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j}$
$\displaystyle-(\mathfrak{b}(n)+C^{\star})\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{3}\sum_{j=1}^{+\infty}\frac{1}{j+3}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{j}\Big{]}\equiv\sum_{k=1}\mathfrak{D}_{k}(z),$

where

$\displaystyle\mathfrak{D}_{1}(z)=C^{\star}\sum_{n=\mathfrak{N}\,+1}^{\infty}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{3}\Big{[}\sum_{j=1}^{+\infty}\frac{3}{j(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{j}+\ln\Big{(}1-\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}\Big{]},$
$\displaystyle\mathfrak{D}_{2}(z)=\sum_{n=\mathfrak{N}\,+1}^{\infty}\Big{[}\frac{\mathfrak{b}(n)z^{3}}{(\mathfrak{b}(n)+C^{\star}+z)^{3}}-\frac{\mathfrak{b}(n)z^{3}}{(\mathfrak{b}(n)+z)^{3}}\Big{]}\Big{[}\sum_{j=1}^{+\infty}\frac{3}{j(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j}+\ln\Big{(}1-\frac{z}{\mathfrak{b}(n)+z}\Big{)}\Big{]},$
$\displaystyle\mathfrak{D}_{3}(z)=\sum_{n=\mathfrak{N}\,+1}^{+\infty}\mathfrak{b}(n)\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{3}\sum_{j=1}^{+\infty}\frac{3z^{j}}{j(j+3)}\Big{(}\frac{1}{(\mathfrak{b}(n)+C^{\star}+z)^{j}}-\frac{1}{(\mathfrak{b}(n)+z)^{j}}\Big{)},$
$\displaystyle\mathfrak{D}_{4}(z)=\sum_{n=\mathfrak{N}\,+1}\mathfrak{b}(n)\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{3}\ln\frac{(\mathfrak{b}(n)+C^{\star})(\mathfrak{b}(n)+z)}{\mathfrak{b}(n)(\mathfrak{b}(n)+C^{\star}+z)}.$

At this point, we treat each term $\mathfrak{D}_{k}(z),$ separately.

$\bullet$ Collecting statement (ii) of Proposition 3.1 with easily verified
relations:

$\displaystyle\Big{|}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{|}<1,\quad\sum_{j=1}^{+\infty}\frac{3}{j(j+3)}\Big{|}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{|}^{j}<\sum_{j=1}^{+\infty}\frac{C}{j^{2}}=C,$
$\displaystyle\Big{|}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\ln\Big{(}1-\frac{z}{\mathfrak{b}(n)+z+C^{\star}}\Big{)}\Big{|}<C,$
(A.5)

with the positive constant $C$ being independent of $n$ and $z$, we end up
with

$|\mathfrak{D}_{1}(z)|\leq C|z|+O(1).$

$\bullet$ Performing simple technical calculations, we conclude

$\mathfrak{D}_{2}(z)=-\sum_{n=\mathfrak{N}\,+1}^{+\infty}\frac{C^{\star}\mathfrak{b}(n)z^{3}[(\mathfrak{b}(n)+z)^{2}+(\mathfrak{b}(n)+z)(\mathfrak{b}(n)+C^{\star}+z)+(\mathfrak{b}(n)+C^{\star}+z)^{2}]}{(\mathfrak{b}(n)+z)^{3}(\mathfrak{b}(n)+C^{\star}+z)^{3}}$
$\times\Big{[}\sum_{j=1}^{+\infty}\frac{3}{j(j+3)}\Big{(}\frac{z}{\mathfrak{b}(n)+z}\Big{)}^{j}+\ln\Big{(}1-\frac{z}{\mathfrak{b}(n)+z}\Big{)}\Big{]}.$

After that, inequalities (Appendix: Proof of Proposition 3.1) and statement
(ii) in Proposition 3.1 arrive at the inequality

$|\mathfrak{D}_{2}(z)|\leq C|z|+O(1).$

$\bullet$ Since $\Big{|}\frac{C^{\star}}{\mathfrak{b}(n)+z}\Big{|}<<1$, we
have

$\displaystyle\frac{1}{(\mathfrak{b}(n)+C^{\star}+z)^{j}}-\frac{1}{(\mathfrak{b}(n)+z)^{j}}$
$\displaystyle=\frac{1}{(\mathfrak{b}(n)+C^{\star}+z)^{j}}\Big{[}1-\Big{(}1+\frac{C^{\star}}{\mathfrak{b}(n)+z}\Big{)}^{j}\Big{]}$
$\displaystyle=\frac{-1}{(\mathfrak{b}(n)+C^{\star}+z)^{j}}\Big{[}\frac{C^{\star}}{\mathfrak{b}(n)+z}+(j-1)O\Big{(}\frac{1}{(\mathfrak{b}(n)+z)^{2}}\Big{)}\Big{]}$

and, accordingly,

$\displaystyle\sum_{j=1}^{\infty}\frac{z^{j}}{j(j+3)}\Big{[}\frac{1}{(\mathfrak{b}(n)+C^{\star}+z)^{j}}-\frac{1}{(\mathfrak{b}(n)+z)^{j}}\Big{]}$
$\displaystyle=\sum_{j=1}^{\infty}\frac{z^{j}}{j(j+3)(\mathfrak{b}(n)+C^{\star}+z)^{j}}\Big{[}\frac{C^{\star}}{\mathfrak{b}(n)+z}+(j-1)O\Big{(}\frac{1}{(\mathfrak{b}(n)+z)^{2}}\Big{)}\Big{]}$
$\displaystyle=\frac{C^{\star}}{\mathfrak{b}(n)+z}\sum_{j=1}^{\infty}\frac{z^{j}}{j(j+3)(\mathfrak{b}(n)+C^{\star}+z)^{j}}\Big{[}1-O\Big{(}\frac{1}{(\mathfrak{b}(n)+z)}\Big{)}\Big{]}$
$\displaystyle+\sum_{j=1}^{\infty}\frac{z^{j}}{(j+3)(\mathfrak{b}(n)+C^{\star}+z)^{j}}O\Big{(}\frac{1}{(\mathfrak{b}(n)+z)^{2}}\Big{)}$
$\displaystyle=\frac{C^{\star}}{\mathfrak{b}(n)+z}\sum_{j=1}^{\infty}\frac{z^{j}}{j(j+3)(\mathfrak{b}(n)+C^{\star}+z)^{j}}\Big{[}1-O\Big{(}\frac{1}{(\mathfrak{b}(n)+z)^{2}}\Big{)}\Big{]}$
$\displaystyle-O\Big{(}\frac{1}{(\mathfrak{b}(n)+z)^{2}}\Big{)}\sum_{j=1}^{\infty}\Big{[}\frac{3z^{j}}{j(j+3)(\mathfrak{b}(n)+C^{\star}+z)^{j}}-\ln\Big{(}1-\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}\Big{]}.$

Finally, coming to $\mathfrak{D}_{3}(z)$ and taking into account (Appendix:
Proof of Proposition 3.1), assumptions in Proposition 3.1 and aforementioned
relations, we deduce

$|\mathfrak{D}_{3}(z)|\leq C|z|+O(1).$

$\bullet$ In order to manage $\mathfrak{D}_{4}(z)$, we take advantage of the
identities

$\ln\frac{(\mathfrak{b}(n)+z)(\mathfrak{b}(n)+C^{\star})}{(\mathfrak{b}(n)+z+C^{\star})\mathfrak{b}(n)}=\ln\frac{\bigg{(}1+\frac{C^{\star}}{\mathfrak{b}(n)}\bigg{)}}{\bigg{(}1+\frac{C^{\star}}{\mathfrak{b}(n)+z}\bigg{)}}=\frac{C^{\star}}{\mathfrak{b}(n)}+O\Big{(}\frac{1}{\mathfrak{b}^{2}(n)}\Big{)}-\frac{C^{\star}}{\mathfrak{b}(n)+z}+O\Big{(}\frac{1}{(\mathfrak{b}(n)+z)^{2}}\Big{)}$

which in turn provide

$\mathfrak{D}_{4}(z)=\sum_{n=\mathfrak{N}\,+1}C^{\star}\Big{[}\mathfrak{b}(n)\\{O((\mathfrak{b}(n)+z)^{-2})+O(\mathfrak{b}^{-2}(n))\\}-\frac{z}{\mathfrak{b}(n)+z}\Big{]}\Big{(}\frac{z}{\mathfrak{b}(n)+C^{\star}+z}\Big{)}^{3}.$

At last, applying statements (i) and (ii) of Proposition 3.1 to the right-hand
side of the last equality, we obtain the desired bound

$|\mathfrak{D}_{4}(z)|\leq C|z|+O(1).$

Collecting representations of all $\mathfrak{D}_{k}$, $k=1,2,3,4,$ we arrive
at the inequality in (iv) of this proposition. Thus, the proof is completed. ∎

## Data Availability Statements.

Data sharing not applicable to this article as no datasets were generated or
analyzed during the current study.

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