text:
\begin{document}

\begin{abstract}
We give an explicit  formula for the discriminant $\Delta_f (x)$ of the  quadrinomials of the form

\[f (x)=x^n+ t (x^2+ax+b)\]

The proof uses  Bezoutians of polynomials.  This paper is an extended version of \cite{o-sh-3}.
\end{abstract}

\maketitle

\section{Introduction}\label{intro}

In \cite{Se} Selmer studied polynomials $f(x)=x^n\pm x \pm 1$ and proved that $x^n-x-1$ is irreducible for all $\geq 2$, while $x^n+x+1$ is irreducible for $n \not\equiv 2 \mod 3$.   In the process he proved that the discriminant of $f(x)$ is (up to a multiplication by $-1$)

$ \Delta = \pm \left( n^n \pm (n-1)^{n-1} \right),$

see \cite{Se} for precise formulas. He also noticed that polynomials $f(x)$ have very small discriminants. He checked his results for $n \leq 20$.   Other mathematicians have considered trinomials $f(x) = x^n + ax+b$ for various reasons; see \cite{Mori}, \cite{Za}.
In general, finding conditions on the coefficients of a polynomial to have minimal discriminant is a difficult problem.  It is related to reduction of binary forms discussed in \cite{reduction} and their heights \cite{height}. More generally it is a special case of  finding conditions on the coefficients of a binary form such that the corresponding  point in the weighted moduli space of invariants is normalized; see \cite{w-height}.

This paper is the first of hopefully others to come to determine for what $a, b, t$ the quadrinomial

\begin{equation}\label{eq-0}
f(x) = x^n + t  (x^2+ax +b)
\end{equation}

has minimal discriminant, is reduced in the sense of \cite{reduction}, or  has minimal naive height.   There have been plenty of efforts to determine such formulas for certain classes of polynomials. In \cite{g-d}, \cite{swa}  the authors focus on  the computation of the discriminant of a trinomial which has been carried out in different ways. In this paper, we determine explicitly a formula for  the discriminant of $f(x)$ by using the approach of Bezoutians.

We prove (see \cref{thm6.0}) that the discriminant $\Delta_f(x)$ of the polynomials in Eq.~\eqref{eq-0} is  given by the formula

\begin{equation}
\Delta =(-1)^{m_1}t^{n-1}   \left( (n-2)^{n-2}(a^2-4b)t^2+\gamma_{\bm c} \, t-n^nb^{n-1} \right),
\end{equation}

such that

\[

\gamma_{\bm c} =  \displaystyle \sum_{k=0}^{m_0}(-1)^{n+k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4} b^k \cdot S_k,

\]

and $m_0=\lfloor (n-3)/2 \rfloor$, $m_1=\lceil (n-3)/2 \rceil$  for $a \neq 0$  or  $a=0$ and $n$ even,  and $\gamma_{\bm c} =0$ for $a=0$ and $n$ odd, and

\begin{small}
\begin{align*}
S_k=&(n-1)^3\binom{n-k-3}{k}a^4-\frac{n(n-1)\left\{ 5n^2-(6k+23)n+10k+24 \right\}}{n-k-3}\binom{n-k-3}{k}a^2b +4n^2(n-2)\binom{n-k-4}{k}b^2.
\end{align*}
\end{small}

It is object of further investigation if such result could be generalized to polynomials $f(x)= x^n+ t \cdot g(x)$, where $g(x)$ is a general cubic.  As a quick application of \cref{thm6.0} we get that   for $b\neq 0$ and $\frac {(n-1)^2 a^2 } {4n (n-2)} \leq b$ the polynomial $f (t, x)$ has no real roots, for any real number $t >0$;  \cref{thm6.1}.   While there is an elementary proof of \cref{thm6.1}, it is interesting to check whether out approach would work for any polynomial of type $f(t, x) = x^n + t \cdot g(x)$, for $\deg g \geq 3$.

It is a quite open problem to determine for what integer values of $a$, $b$, and  $t$ the quadrinomial $f(x)$ is irreducible or the discriminant $\Delta_f$ has a minimal value.  Moreover, it is our intention to study in the future for what conditions on $a, b, t$ the quadrinomial is reduced in the sense of \cite{reduction}.

Another motivation of looking at the  discriminant of such family of polynomials comes from our efforts to construct superelliptic Jacobians with large endomorphism rings.  We have checked computationally that for small $n$ and  $f(t, x) \in \Q[t, x]$, for almost all $t$ the Galois group $\Gal_{\Q} (f, x)$ is isomorphic to $S_n$.  Due to results of Zarhin this implies that the superelliptic curve $y^m = f(t, x)$ has large endomorphism ring; see \cite{f-sh} for related matters. Hence, curves $y^n=f(x)$, where $f(x)$   is as above, are smooth curves whose  Jacobians are expected to have large endomorphism rings. This remains the focus of further investigation.

\bigskip

\noindent \textbf{Notation:} Throughout this paper $n$ is a positive integer,  $\Delta$   denotes the discriminant of a polynomial $f(x)$ with respect to the variable $x$.
The symbol   $\lfloor \cdot \rfloor$ is the floor function and $\lceil \cdot \rceil$ is the ceiling function.  $B_n (f, g)$ denotes the level $n$ Bezoutian of two polynomials $f$ and $g$. For a polynomial $f \in \R[x]$ the symbol $N_f$ denotes the number of real roots of $f(x)$.  We denote by $\binom{n}{m}$ $(n,m \in \Z_{\geq 0})$ the binomial coefficient.

\section{Preliminaries}

Let $F$ be a field of characteristic zero and $f_1(x)$, $f_2(x)$ be polynomials over $F$. Then, for any integer $n$ such that  $n \geq \max \{ \mathrm{deg} f_1, \mathrm{deg} f_2 \}$, we put

\begin{align*}
B_{n}(f_{1},f_{2}) :&= \frac{f_{1}(x)f_{2}(y)-f_{1}(y)f_{2}(x)}{x-y}= \sum_{i,j=1}^{n} \alpha _{ij}x^{n-i}y^{n-j}\in F[x,y], \\
M_{n}(f_{1},f_{2}) :&= (\alpha_{ij})_{1 \leq i,j \leq n}.
\end{align*}
The $n \times n$ matrix $M_{n}(f_{1},f_{2})$ is called the Bezoutian of $f_1$ and $f_2$. When $f_2=f_1^{\prime}$, the formal derivative of $f_1$ with respect to the indeterminate $x$, we often write $B_n(f_1):=B_{n}(f_{1},f_{1}^{\prime})$, $M_n(f_1):=M_n(f_1, f_1^{\prime})$ and the matrix $M_n(f_1)$ is called the Bezoutian of $f_1$. We denote by $N_{f_1}$ the number of distinct real roots of $f_1(x)$ and $\Delta(f_1)$ the discriminant of $f_1(x)$. Moreover, for any real  symmetric matrix $M$, we denote by $\sigma(M)$ the index of inertia of $M$. The following are the list of important properties of Bezoutians.

\begin{lem}\label{prop2.1}
Notations as above, we have
\begin{enumerate}
\item[$(i)$] $M_n(f_1, f_2)$ is an $n \times n$ symmetric matrix over $F$. $(M_n(f_1, f_2) \in \sym_n(F))$.
\item[$(ii)$] $B_{n}(f_{1},f_{2})$ $(M_{n}(f_{1},f_{2}))$ is linear in $f_{1}$ and $f_{2}$, separately.
\item[$(iii)$] $B_{n}(f_{1},f_{2})=-B_{n}(f_{2},f_{1})$ $(M_{n}(f_{1},f_{2})=-M_{n}(f_{2},f_{1}))$.
\item[$(iv)$] $N_{f_1}=\sigma(M_n(f_1))$.
\item[$(v)$] $\displaystyle \Delta(f_1)=\frac{1}{led(f_1)^2}\det M_n(f_1)$, where $led(f_1)$ is the leading coefficient of $f_1$. In particular, if $f_1$ is monic, we have $\displaystyle \Delta(f_1)=\det M_n(f_1)$.
\item[$(vi)$] Let $\lambda ,\mu ,\nu$ be integers such that $\lambda \geq \mu > \nu \geq 0$. Then
$M_{\lambda }(x^{\mu }, x^{\nu }) = (m_{ij} )_{1\leq i,j\leq \lambda }$,
where
\begin{equation*}
m_{ij}=\begin{cases}
               1 & \text{$i+j=2\lambda-(\mu+\nu)+1$ \ $(\lambda-\mu+1\leq i,j\leq \lambda-\nu )$}, \\
               0 & \text{otherwise}.
             \end{cases}
\end{equation*}
\end{enumerate}
\end{lem}

\begin{proof}
For (i)--(iv), see \cite[Theorem 8.25, 9.2]{fuh}. For (v), see equation (4) of \cite[p.217]{fro}. For (vi), see \cite[Lemma 2]{o-sh}
\end{proof}
For any vector ${\bm r}=(r_0, \cdots, r_s) \in \R^{s+1}$, let us put
\begin{align*}
g_{\bm r}(x)=g(r_0,\cdots, r_s ; x) = \sum_{k=0}^{s}r_{s-k}x^{s-k} \in \R[x], \\
f_{\bm r}(t; x)=f^{(n)}(r_0,\cdots, r_s, t ; x)=x^n+t \cdot g_{\bm r}(x) \in \R(t)[x].
\end{align*}

In the previous paper \cite{o-sh}, by using the above properties of Bezoutians, we obtained the next theorem

\begin{thm}{\cite[Thm.~2]{o-sh}} \label{thm5.2}
Let ${\bm r}=(r_0, \cdots, r_s) \in \R^{s+1}$ be a vector such that $g_{\bm r}(x)$ is a degree $s$ separable polynomial satisfying $N_{g_{\bm r}(x)}=\gamma$ $(0 \leq \gamma \leq s)$. Let us consider $f_{\bm r}(t ; x)=f^{(n)}(r_0, \cdots, r_s, t ; x)$ as a polynomial over $\R(t)$ in $x$ and put
\[
P_{\bm r}(t)=\det M_n(f_{\bm r}(t ; x))=\det M_n(f_{\bm r}(t ; x), f_{\bm r}^{\prime}(t ; x)),
\]
where $f_{\bm r}^{\prime}(t ; x)$ is a derivative of $f_{\bm r}(t ; x)$ with respect to $x$. Then, for any real number $t>\alpha_{\bm r}=\max\{ \alpha \in \R \mid P_{\bm r}(\alpha)=0 \}$, we have
\begin{align*}
N_{f_{\bm r}(t ; x)}
=
\begin{cases}
\gamma+1 & \text{$n-s$ $:$ odd} \\
\gamma & \text{$n-s$ $:$ even, $r_s>0$} \\
\gamma+2 & \text{$n-s$ $:$ even, $r_s<0$}.
\end{cases}
\end{align*}
\end{thm}

Then, by using \cref{thm5.2} and  \cref{thm6.0}, we construct a certain family of totally complex polynomials of the form $f_{(b,a,1)}(t; x)$; see \cref{thm6.1}. Here, note that a real polynomial $f(x)$ to be called totally complex if it has no real roots, that is, $N_f=0$.

Let $n \geq 3$ be an integer and $a$, $b$ be real numbers.   Let us put ${\bm c}=(b,a,1) \in \R^3$ and

\begin{equation*}
g_{\bm c}(x)=x^2+ax+b \in \R[x], \ f_{\bm c}(t; x)=x^n+tg_{\bm c}(x) \in \R(t)[x].
\end{equation*}

We denote by $\binom{n}{m}$ $(n,m \in \Z_{\geq 0})$ the binomial coefficient. Note that $\binom{n}{0}=1$ for any integer $n \geq 0$ and $\binom{n}{m}=0$ if $n<m$. Moreover, we define $\binom{n}{m}=0$ for any $n \in \Z_{\geq 0}$ and $m \in \Z_{<0}$.

Our main result is the following.

\begin{thm} \label{thm6.0}
Put $m_0=\lfloor (n-3)/2 \rfloor$, $m_1=\lceil (n-3)/2 \rceil$. Moreover, for any integers $n\geq 4$ and $k$ $\left(0 \leq k \leq m_0\right)$, put

\begin{small}
\begin{align*}
S_k=&(n-1)^3\binom{n-k-3}{k}a^4-\frac{n(n-1)\left\{ 5n^2-(6k+23)n+10k+24 \right\}}{n-k-3}\binom{n-k-3}{k}a^2b +4n^2(n-2)\binom{n-k-4}{k}b^2.
\end{align*}
\end{small}

Here, $\lfloor \cdot \rfloor$ is the floor function and $\lceil \cdot \rceil$ is the ceiling function. \\[1mm]
$(1)$ \ Suppose $n=3$. Then, we have

\begin{align*}
\Delta\left(f_{\bm c}(t; x)\right)=t^2\Biggl\{(a^2-4b)t^2-(4a^3-18ab)t-27b^2 \Biggr\}.
\end{align*}

$(2)$ \ Suppose $n\geq 4$. Then, we have

\begin{align*}
\Delta\left(f_{\bm c}(t; x)\right)=(-1)^{m_1}t^{n-1}\Biggl\{(n-2)^{n-2}(a^2-4b)t^2+\gamma_{\bm c}t-n^nb^{n-1} \Biggr\},
\end{align*}
where

\begin{align*}
\gamma_{\bm c}=
\begin{cases}
\displaystyle \sum_{k=0}^{m_0}(-1)^{n+k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^kS_k & \hspace{-1mm} \text{$(a \neq 0$ or $a=0$, $n:$ even$)$}, \vspace{2mm} \\
0 & \hspace{-1mm} \text{$(a=0$, $n:$ odd$)$}.
\end{cases}
\end{align*}
\end{thm}

As a corollary of the above we get.

\begin{cor} \label{thm6.1}
Suppose $n\geq 4$ is an even integer. Moreover, let us suppose $b \neq 0$ and $(n-1)^2a^2/4n(n-2)\leq b$. Then we have $N_{f_{\bm c}(t; x)}=0$ for any positive real number $t$.
\end{cor}

By a direct computation, we have
\begin{align*}
&\Delta\left(x^3+t(x^2+ax+b)\right)=t^2\left\{ (a^2-4b)t^2-(4a^3-18ab)t-27b^2 \right\}, \\
&\Delta\left(x^4+t(x^2+ax+b)\right)=-t^3\left\{ (4a^2-16b)t^2+(27a^4-144a^2b+128b^2)t-256b^3 \right\}
\end{align*}
and we get \cref{thm6.0} for $n=3$, $4$. Moreover, if $n=4$ and $(n-1)^2a^2/4n(n-2)=b$, we have
\begin{align*}
P_{\bm c}(t)=\Delta\left(x^4+t(x^2+ax+b)\right)=\frac{1}{2}a^2\left(t+\frac{27}{8}a^2 \right)^2t^3
\end{align*}
and hence $\alpha_{\bm c}=0$, which implies $N_{f_{\bm c}(t; x)}=0$ for any $t>0$ by \cref{thm5.2} since $x^2+ax+b$ has no real root in this case. Then, by considering the graph of the function $y=x^4+t(x^2+ax+b)$, we have $N_{f_{\bm c}(t; x)}=0$ whenever $(n-1)^2a^2/4n(n-2)\leq b$ and $t>0$, which is the claim of \cref{thm6.1} for $n=4$. Therefore, let us assume $n \geq 5$ hereafter.

In the following, we put
\begin{align*}
&A_{\bm c}(t)=(a_{ij}^{(\bm c)}(t))_{1 \leq i, j \leq n}=M_n(f_{\bm c}(t; x)) \in \mathrm{Sym}_n(\R(t)), \\
&B_{\bm c}=(b_{ij}^{(\bm c)})_{1 \leq i,j \leq 2}=M_2(g_{\bm c}(x)) \in \mathrm{Sym}_{2}(\R), \\
&P_{\bm c}(t)=\det A_{\bm c}(t)=\Delta(f_{\bm c}(t; x)).
\end{align*}
Then, by   \cref{thm5.2}, we have the next lemma.

\begin{cor} \label{lem6.1}
Suppose $n$ is even and $a^2/4<b$. Put $\alpha_{\bm c}=\max \{ \alpha \in \R \mid P_{\bm c}(\alpha)=0 \}$. Then, for any real number $t>\alpha_{\bm c}$, we have $N_{f_{\bm c}(t, x)}=0$.
\end{cor}

To prove \cref{thm6.0}, we will use the equality $\Delta(f_{\bm c}(t; x))=\det A_{\bm c}(t)$. Also, to prove \cref{thm6.1}, it is enough to prove $\alpha_{\bm c}=0$ when $(n-1)^2a^2/4n(n-2)\leq b$ and $(a,b)\neq(0,0)$, which implies we need to compute the polynomial $P_{\bm c}(t)=\det A_{\bm c}(t)$ precisely. Here, let $Q_{m}(k;c)=(q _{ij})_{1\leq i,j\leq m}$,  $R_{m}(k,l;c)=(r _{ij})_{1\leq i,j\leq m}$ be $m\times m$ elementary matrices such that
\begin{center}
\noindent $Q_m(k;c)$=
\scalebox{0.85}[1]
{$\left[
\begin{array}{ccccccc}
1 & & & & & & \\
& \ddots & & & & & \\
& & 1 & & & & \\
& & & c & & & \\
& & & & 1 & & \\
& & & & & \ddots & \\
& & & & & & 1
\end{array}
\right]
$},
$R_m(k,l;c)$=
\scalebox{0.85}[1]
{$\left[
\begin{array}{cccccccc}
1 & & & & & & & \\
& \ddots & & & & & & \\
& & 1 & & & c & \\
& & & \ddots & & & & \\
& & & & & 1 & & \\
& & & & & & \ddots & \\
& & & & & & & 1
\end{array}
\right],
$}
\end{center}

where $q _{kk}=c$, $r _{kl}=c$, respectively and, for any $m \times m$ matrices $M_1$, $M_2$, $\cdots$, $M_l$, put
$\prod_{k=1}^{l} M_k=M_1M_2\cdots M_l$. Moreover, put $l_k=n+k$. Then, as is the case in \cite{o-sh}, we inductively define the matrix $A_{\bm c}(t)_k$ $(1 \leq k\leq n-2)$ as follows:

\begin{itemize}

\item[(1)] $A_{\bm c}(t)_1=(a_{ij}^{(\bm c)}(t)_1)_{1 \leq i,j \leq n}={}^tS_{\bm c}(t)_1A_{\bm c}(t)S_{\bm c}(t)_1$, where

\[
S_{\bm c}(t)_1=\displaystyle Q_{n}(1;1/\sqrt{n})\prod_{k=0}^{1}R_n(1,l_k-1;-a_{1, l_k-1}^{(\bm c)}(t)/\sqrt{n}).
\]

\item[(2)] Put

\[
n_0 =
\begin{cases}
(n-1)/2, & \text{$n$ $:$ odd}, \\
n/2, & \text{$n$ $:$ even}.
\end{cases}
\]

Then, $A_{\bm c}(t)_{k}=(a_{ij}^{(\bm c)}(t)_k)_{1 \leq i,j \leq n}={}^tS_{\bm c}(t)_kA_{\bm c}(t)_{k-1}S_{\bm c}(t)_k$, where

\[
S_{\bm c}(t)_k=
\begin{cases}
\displaystyle \prod_{m=l_0-k+1}^{n}R_n\left(l_0-k,m;-\dfrac{a_{km}^{({\bm c})}(t)_{k-1}}{-(n-2)t}\right), &     \hspace{-24mm}\text{($2 \leq k \leq n_0$)}  \\[3mm]
\displaystyle R_n\left(l_0-k,k;-\dfrac{a_{kk}^{({\bm c})}(t)_{k-1}}{-2(n-2)t}\right)\prod_{m=k+1}^{n}R_n\left(l_0-k,m;-\dfrac{a_{km}^{({\bm c})}(t)_{k-1}}{-(n-2)t}\right), \; \;  \text{($n_0 < k \leq n-2$)}.
\end{cases}
\]

\end{itemize}

By the definition of $A_{\bm c}(t)_k$ $(1 \leq k\leq n-2)$, we have $\det A_{\bm c}(t)=n\det A_{\bm c}(t)_{n-2}$. Therefore, to compute the polynomial $P_{\bm c}(t)=\det A_{\bm c}(t)$, let us first compute the matrix $A_{\bm c}(t)_{n-2}$ concretely.

\section{Computation of the matrix $A_{\mbox{\scriptsize \boldmath $c$}}(t)_{n-2}$}

By \cref{prop2.1}, we have

\[
B_{\bm c} =M_2(x^2+ax+b, 2x+a) =2M_2(x^2, x)+aM_2(x^2, 1)+(a^2-2b)M_2(x,1) =
\left[
\begin{array}{cc}
2 & a \\
a & a^2-2b
\end{array}
\right].
\]

Here, let us give some examples of the matrices $A_{\bm c}(t)$ and $A_{\bm c}(t)_1$ for some small $n$.

\begin{exa} \label{exa6.1}

\begin{enumerate}

\item
Let $n=5$. Then,
\begin{align*}
A_{\bm c}(t)&=
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{ccccc}
5 & 0 & 0 & 2t & at \\
0 & 0 & -3t & -4at & -5bt \\
0 & -3t & -4at & -5bt & 0 \\
2t & -4at & -5bt & 2t^2 & at^2 \\
at & -5bt & 0 & at^2 & (a^2-2b)t^2
\end{array}
\right]
$}, \quad
A_{\bm c}(t)_1&=
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & -3t & -4at & -5bt \\
0 & -3t & -4at & -5bt & 0 \\
0 & -4at & -5bt & (6/5)t^2 & (3/5)at^2 \\
0 & -5bt & 0 & (3/5)at^2 & \left\{(4/5)a^2-2b\right\}t^2
\end{array}
\right]
$}.
\end{align*}

\item
Let $n=6$. Then,

\begin{small}
\[
A_{\bm c}(t) =
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{cccccc}
6 & 0 & 0 & 0 & 2t & at \\
0 & 0 & 0 & -4t & -5at & -6bt \\
0 & 0 & -4t & -5at & -6bt & 0 \\
0 & -4t & -5at & -6bt & 0 & 0 \\
2t & -5at & -6bt & 0 & 2t^2 & at^2 \\
at & -6bt & 0 & 0 & at^2 & (a^2-2b)t^2
\end{array}
\right]
$}, \;
A_{\bm c}(t)_1 =
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -4t & -5at & -6bt \\
0 & 0 & -4t & -5at & -6bt & 0 \\
0 & -4t & -5at & -6bt & 0 & 0 \\
0 & -5at & -6bt & 0 & \frac 4 3 t^2 & \frac 2 3 at^2 \\
0 & -6bt & 0 & 0 & \frac 2 3 at^2 & \left( \frac 5 6 a^2-2b \right) t^2
\end{array}
\right]
$}.
\]
\end{small}

\end{enumerate}
\qed
\end{exa}

Let's continue our discussion for  $n\geq 5$. Then, by \cite[Prop~4]{o-sh} and  \cite[Eq.~(5)]{o-sh}, we have

\begin{align} \label{eq6.1}
A_{\bm c}(t)
=
\scalebox{0.81}[0.81]
{$\left[
\begin{array}{cccccccc}
n & 0 & \dots & \dots & \dots & 0 & 2t & at \\
0 &&&&& -(n-2)t & -(n-1)at & -nbt \\
\vdots &&&&  \rotatebox[origin=c]{292}{\vdots} &  -(n-1)at & -nbt & 0 \\
\vdots &&& \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & -nbt & 0 & \vdots \\
\vdots && \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \vdots & \vdots \\
0 & -(n-2)t & -(n-1)at & -nbt & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} && 0 & 0 \\
2t & -(n-1)at & -nbt & 0 & \dots & 0 & 2t^2 & at^2 \\
at & -nbt & 0 & \dots & \dots & 0 & at^2 & (a^2-2b)t^2
\end{array}
\hspace{-0.5mm}\right]
$}
\end{align}

and hence

\begin{align} \label{eq6.2}
A_{\bm c}(t)_1
=
\scalebox{0.82}[0.82]
{$\left[
\begin{array}{cccccccc}
1 & 0 & \dots & \dots & \dots & 0 & 0 & 0 \\
0 &&&&& -(n-2)t & -(n-1)at & -nbt \\
\vdots &&&&  \rotatebox[origin=c]{292}{\vdots} &  -(n-1)at & -nbt & 0 \\
\vdots &&& \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & -nbt & 0 & \vdots \\
\vdots && \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \vdots & \vdots \\
0 & -(n-2)t & -(n-1)at & -nbt & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} && 0 & 0 \\
0 & -(n-1)at & -nbt & 0 & \dots & 0 & 2\left(1-\frac{2}{n}\right)t^2 &  \left(1-\frac{2}{n}\right)at^2 \\
0 & -nbt & 0 & \dots & \dots & 0 & \left(1-\frac{2}{n}\right)at^2 & \left\{\left(1-\frac{1}{n}\right)a^2-2b\right\}t^2
\end{array}
\right].
$}
\end{align}

Here, similar to $A_{\bm c}(t)_{k}$ $(2 \leq k \leq n-2)$, we inductively define the matrix $W(t)_{k}=(w_{ij}(t)_k)_{1\leq i,j \leq n}$ $(2 \leq k \leq n-2)$ as follows; \\[3mm]

(1) $W(t)_1=(w_{ij}(t)_1)_{1 \leq i,j \leq n}$, where
\begin{align*}
w_{ij}(t)_1
=
\begin{cases}
qt, & \text{$i+j=n$, $2 \leq i, j \leq n-2$}, \\
rt, & \text{$i+j=n+1$, $2 \leq i, j \leq n-1$}, \\
st, & \text{$i+j=n+2$, $2 \leq i, j \leq n$}, \\
0, & \text{otherwise}.
\end{cases}
\end{align*}

(2)  $W(t)_{k}=(w_{ij}(t)_k)_{1\leq i,j \leq n}$ $(2 \leq k \leq n-2)={}^tS(t)_kW(t)_{k-1}S(t)_k$, where

\begin{align*}
S(t)_k=
\begin{cases}
\displaystyle \prod_{m=n-k+1}^{n}R_n\left(n-k,m;-\dfrac{w_{km}(t)_{k-1}}{qt}\right) &\hspace{-24mm}\text{($2 \leq k \leq n_0$)}, \\[3mm]
\displaystyle R_n\left(n-k,k;-\dfrac{w_{kk}(t)_{k-1}}{-2qt}\right)\prod_{m=k+1}^{n}R_n\left(n-k,m;-\dfrac{w_{km}(t)_{k-1}}{qt}\right), \hfill \text{($n_0 < k \leq n-2$)}.
\end{cases}
\end{align*}

\begin{exa}\label{exa3}

\begin{enumerate}
\item Let $n=5$. Then,

\begin{align*}
W(t)_1 =
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{ccccc}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & qt & rt & st \\
0 & qt & rt & st & 0 \\
0 & rt & st & 0 & 0 \\
0 & st & 0 & 0 & 0
\end{array}
\right]
$}, \; \;

& W(t)_{n_0}=
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{ccccc}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & qt & 0 & 0 \\
0 & qt & rt & (qs-r^2)t/q & -rst/q \\
0 & 0 & (qs-r^2)t/q & -(2qs-r^2)rt/q^2 & -(qs-r^2)st/q^2 \\
0 & 0 & -rst/q & -(qs-r^2)st/q^2 & rs^2t/q^2
\end{array}
\right]
$}, \\

W(t)_{n-2}&=
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{ccccc}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & qt & 0 & 0 \\
0 & qt & 0 & 0 & 0 \\
0 & 0 & 0 & -(2qs-r^2)rt/q^2 & -(qs-r^2)st/q^2 \\
0 & 0 & 0 & -(qs-r^2)st/q^2 & rs^2t/q^2
\end{array}
\right]
$}.
\end{align*}

\item
Let $n=6$. Then,
\begin{align*}
W(t)_1&=
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{cccccc}
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & qt & rt & st \\
0 & 0 & qt & rt & st & 0 \\
0 & qt & rt & st & 0 & 0 \\
0 & rt & st & 0 & 0 & 0 \\
0 & st & 0 & 0 & 0 & 0
\end{array}
\right]
$}, \\
W(t)_{n_0}&=
\scalebox{0.88}[0.9]
{$\left[
\begin{array}{cccccc}
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & qt & 0 & 0 \\
0 & 0 & qt & 0 & 0 & 0 \\
0 & qt & 0 & (qs-r^2)t/q & -(2qs-r^2)rt/q^2 & -(qs-r^2)st/q^2 \\
0 & 0 & 0 & -(2qs-r^2)rt/q^2 & -(q^2s^2-3qr^2s+r^4)t/q^3 & (2qs-r^2)rst/q^3 \\
0 & 0 & 0 & -(qs-r^2)st/q^2 & (2qs-r^2)rst/q^3 & (qs-r^2)s^2t/q^3
\end{array}
\right]
$}, \\
W(t)_{n-2}&=
\scalebox{0.9}[0.9]
{$\left[
\begin{array}{cccccc}
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & qt & 0 & 0 \\
0 & 0 & qt & 0 & 0 & 0 \\
0 & qt & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -(q^2s^2-3qr^2s+r^4)t/q^3 & (2qs-r^2)rst/q^3 \\
0 & 0 & 0 & 0 & (2qs-r^2)rst/q^3 & (qs-r^2)s^2t/q^3
\end{array}
\right]
$}.
\end{align*}
\end{enumerate}
\end{exa}

In the same way, let us compute the matrix $W(t)_{n-2}$ for any integer $n\geq 7$.

\begin{lem} \label{lem6.2}
Suppose $n\geq 7$. Let $\{ x_m \}$, $\{ y_m \}$ be sequences defined by the next recurrence relations;
\begin{align*}
&x_0=0, \ x_1=-qt, \ x_2=rt, \ x_{m+2}=-\frac{r}{q}x_{m+1}-\frac{s}{q}x_m \ (m\geq 1), \\
&y_0=y_1=0, \ y_{m+1}=-\frac{s}{q}x_{m} \ (m \geq 1)
\end{align*}
and put
\begin{align*}
n_1=
\begin{cases}
n_0-1 & \text{$n:$ odd}, \\
n_0-2 & \text{$n:$ even}.
\end{cases}
\end{align*}
Then, for any integer $k$ such that $2 \leq k \leq n_1$, we have
\begin{align*}
w_{ij}(t)_{k}
=
\begin{cases}
0 & \text{$(i,j)=(k, \ell)$ or $(\ell, k)$ $(n-k+1\leq \ell \leq n)$}, \\
x_{\ell-n+k+2} & \text{$(i,j)=(k+1, \ell)$ or $(\ell, k+1)$ $(n-k \leq \ell \leq n-1)$}, \\
y_{\ell-n+k+2} & \text{$(i,j)=(k+2, \ell)$ or $(\ell, k+2)$ $(n-k \leq \ell \leq n-1)$}, \\
- \frac {sx_{k}} q & \text{$(i,j)=(k+1, n)$ or $(n, k+1)$}, \\
- \frac {sy_{k}} q & \text{$(i,j)=(k+2, n)$ or $(n, k+2)$}, \\
w_{ij}(t)_{k-1} & \text{otherwise}.
\end{cases}
\end{align*}
\end{lem}

\begin{proof}
Let us prove this lemma by induction on $k$. First, suppose $k=2$. Then, by the definition of $W(t)_2$, we have

\begin{align*}
W(t)_2
=
\scalebox{0.76}[1]
{$\left[
\begin{array}{cccccccccc}
0 & \cdots & \cdots & \dots & \dots & \dots & 0 & 0 & 0 & 0 \\
\vdots &&&&&&& qt & 0 & 0 \\
\vdots &&&&&& qt & rt & \frac {(qs-r^2)t} q & - \frac {rst} q \\
\vdots &&&&& \rotatebox[origin=c]{300}{\vdots} & rt & st & - \frac {rst} q & - \frac {s^2t} q \\
\vdots &&&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & st & 0 & 0 & 0 \\
\vdots &&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \vdots \\
0 & 0 & qt & rt & st & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} && \vdots \\
0 & qt & rt & st & 0 & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} &&& \vdots \\
0 & 0 & (qs-r^2) \frac t q & - \frac {rst} q & 0 & \rotatebox[origin=c]{300}{\vdots} &&&& \vdots \\
0 & 0 & - \frac {rst} q & - \frac {s^2t} q & 0 & \cdots & \cdots & \cdots & \cdots & 0
\end{array}
\right],
$}
\end{align*}

which implies the claim of  \cref{lem6.2} for $k=2$ since

\begin{align*}
x_2=rt, \ x_3=\frac{(qs-r^2)t}{q}, \ y_2=st, \ y_3=-\frac{rst}{q}.
\end{align*}

Next, suppose  \cref{lem6.2} is true for $k=2, \cdots, m-1$ $(m-1<n_1)$. Then, since

\begin{align*}
W(t)_{m-1}
=
\scalebox{0.70}[0.7]
{$\left[
\begin{array}{ccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&&&&&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \vdots & \vdots \\
\vdots &&&&&&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&&&&&& qt & x_2 & x_3 & \cdots & x_{m} & -sx_{m-1}/q \\
\vdots &&&&&& \rotatebox[origin=c]{300}{\vdots} & rt & y_2 & y_3 & \cdots & y_{m} & -sy_{m-1}/q \\
\vdots &&&&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & st & 0 & 0 & \cdots & 0 & 0 \\
\vdots &&&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} &&& \rotatebox[origin=c]{300}{\vdots} & \vdots \\
\vdots &&& qt & rt & st & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} &&& \rotatebox[origin=c]{300}{\vdots} && \vdots \\
\vdots && qt & x_2 & y_2 & 0 & \rotatebox[origin=c]{300}{\vdots} &&& \rotatebox[origin=c]{300}{\vdots} &&& \vdots \\
\vdots & \rotatebox[origin=l]{300}{\vdots} & 0 & x_3 & y_3 & 0 &&& \rotatebox[origin=c]{300}{\vdots} &&&& \vdots \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & \vdots & \vdots & \vdots & \vdots && \rotatebox[origin=c]{300}{\vdots} &&&&& \vdots \\
0 & \cdots & 0 & x_{m} & y_{m} & 0 & \rotatebox[origin=c]{300}{\vdots} &&&&&& \vdots \\
0 & \cdots & 0 & -sx_{m-1}/q & -sy_{m-1}/q & 0 & \cdots & \cdots& \cdots & \cdots & \cdots & \cdots & 0
\end{array}
\right],
$}
\end{align*}
\normalsize
we have

\begin{align*}
W(t)_{m}
=
\scalebox{0.72}[0.72]
{$\left[
\begin{array}{ccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&&&&&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \vdots & \vdots \\
\vdots &&&&&&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&&&&&& qt & 0 & 0 & \cdots & 0 & 0 \\
\vdots &&&&&& \rotatebox[origin=c]{300}{\vdots} & rt & x_2^{\prime} & x_3^{\prime} & \cdots & x_{m}^{\prime} & x_{m+1}^{\prime} \\
\vdots &&&&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & st & y_2^{\prime} & y_3^{\prime} & \cdots & y_{m}^{\prime} & y_{m+1}^{\prime} \\
\vdots &&&& \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} &&& \rotatebox[origin=c]{300}{\vdots} & \vdots \\
\vdots &&& qt & rt & st & \rotatebox[origin=c]{300}{\vdots} & \rotatebox[origin=c]{300}{\vdots} &&& \rotatebox[origin=c]{300}{\vdots} && \vdots \\
\vdots && qt & 0 & x_2^{\prime} & y_2^{\prime} & \rotatebox[origin=c]{300}{\vdots} &&& \rotatebox[origin=c]{300}{\vdots} &&& \vdots \\
\vdots & \rotatebox[origin=l]{300}{\vdots} & 0 & 0 & x_3^{\prime} & y_{3}^{\prime} &&& \rotatebox[origin=c]{300}{\vdots} &&&& \vdots \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & \vdots & \vdots & \vdots & \vdots && \rotatebox[origin=c]{300}{\vdots} &&&&& \vdots \\
0 & \cdots & 0 & 0 & x_{m}^{\prime} & y_{m}^{\prime} & \rotatebox[origin=c]{300}{\vdots} &&&&&& \vdots \\
0 & \cdots & 0 & 0 & x_{m+1}^{\prime} & y_{m+1}^{\prime} & \cdots & \cdots& \cdots & \cdots & \cdots & \cdots & 0
\end{array}
\right],
$}
\end{align*}
\normalsize
where
\begin{align*}
&x_{\ell}^{\prime}=y_{\ell}-\frac{x_{\ell}}{qt}\cdot rt=-\frac{r}{q}x_{\ell}-\frac{s}{q}x_{\ell-1}=x_{\ell+1} \ (2 \leq \ell \leq m), \\
&y_{\ell}^{\prime}=-\frac{x_{\ell}}{qt}\cdot st=-\frac{s}{q}x_{\ell}=y_{{\ell}+1} \ (2 \leq \ell \leq m), \\
&x_{m+1}^{\prime}=-\frac{s}{q}y_{m-1}+\left(-\frac{r}{q}\right)\cdot \left(-\frac{s}{q}x_{m-1}\right)=-\frac{s}{q}\left( -\frac{r}{q}x_{m-1}-\frac{s}{q}x_{m-2} \right)=-\frac{s}{q}x_{m} \\
&y_{m+1}^{\prime}=\left(-\frac{s}{q}\right)\cdot \left(-\frac{s}{q}x_{m-1}\right)=-\frac{s}{q}y_{m}.
\end{align*}
This completes the proof of  \cref{lem6.2}.
\end{proof}

\begin{prop} \label{lem6.4}
For any integer $n\geq 5$, we have

\begin{align*}
w_{ij}(t)_{n-2}
=
\begin{cases}
qt & \text{$i+j=n$ $(2 \leq i,j \leq n-2)$}, \\
x_{n-1} & \text{$(i,j)=(n-1,n-1)$}, \\
y_{n-1} & \text{$(i,j)=(n-1,n)$ or $(n,n-1)$}, \\
(-s/q)y_{n-2} & \text{$(i,j)=(n,n)$}, \\
0 & \text{otherwise}.
\end{cases}
\end{align*}
\end{prop}

\begin{proof}
 \cref{lem6.4} has been proved for $n=5$, $6$ in  \cref{exa3}. Thus, we suppose $n\geq 7$. First, by solving the recurrence relation given in  \cref{lem6.2}, we have

\begin{align} \label{eq6.3}
x_m=
\begin{cases}
\frac{(-1)^mmr^{m-1}}{2^{m-1}q^{m-2}}t & \text{$(r^2-4qs=0)$}, \\[4mm]
\frac{(-1)^m\left\{ \left(r+\sqrt{r^2-4qs}\right)^m-\left(r-\sqrt{r^2-4qs}\right)^m \right\}}{2^mq^{m-2}\sqrt{r^2-4qs}}t & \text{$(r^2-4qs \neq 0)$}.
\end{cases}
\end{align}

Note that

\begin{align*}
&\frac{(-1)^m\left\{ \left(r+\sqrt{r^2-4qs}\right)^m-\left(r-\sqrt{r^2-4qs}\right)^m \right\}}{2^mq^{m-2}\sqrt{r^2-4qs}}t =\frac{(-1)^mt}{2^mq^{m-2}}\cdot2\sum_{k=0}^{\lfloor \frac {m-1} 2 \rfloor}\binom{m}{2k+1}r^{m-2k-1}(r^2-4qs)^k
\end{align*}

and hence, by putting $r^2-4qs=0$, we have

\begin{align*}
\frac{(-1)^m\left\{ \left(r+\sqrt{r^2-4qs}\right)^m-\left(r-\sqrt{r^2-4qs}\right)^m \right\}}{2^mq^{m-2}\sqrt{r^2-4qs}}t=\frac{(-1)^mmr^{m-1}}{2^{m-1}q^{m-2}}t,
\end{align*}

the solution for the case of $r^2-4qs=0$. Here, suppose $n$ is odd. Then, we have $n_1= \frac {n-3} 2$ and hence by  \cref{lem6.2},

\begin{align*}
w_{ij}(t)_{n_1}
=
\begin{cases}
0 & \text{$(i,j)=((n-3)/2, \ell)$ or $(\ell, (n-3)/2)$ $((n+5)/2\leq \ell \leq n)$}, \\
x_{\ell-(n-1)/2} & \text{$(i,j)=((n-1)/2, \ell)$ or $(\ell, (n-1)/2)$ $((n+3)/2 \leq \ell \leq n-1)$}, \\
y_{\ell-(n-1)/2} & \text{$(i,j)=((n+1)/2, \ell)$ or $(\ell, (n+1)/2)$ $((n+3)/2 \leq \ell \leq n-1)$}, \\
-sx_{(n-3)/2}/q & \text{$(i,j)=((n-1)/2, n)$ or $(n, (n-1)/2)$}, \\
-sy_{(n-3)/2}/q & \text{$(i,j)=((n+1)/2, n)$ or $(n, (n+1)/2)$}, \\
w_{ij}(t)_{(n-5)/2} & \text{otherwise}.
\end{cases}
\end{align*}

Therefore, we have the next expression of the matrix $W(t)_{n_1}$;

\[
W(t)_{n_1}
=
\scalebox{0.75}[0.75]
{$\left[
\begin{array}{ccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&&& \rotatebox[origin=c]{300}{\vdots} && \vdots & \vdots \\
\vdots &&&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&&& qt & x_2 & x_3 & \cdots & x_{(n-1)/2} & -sx_{(n-3)/2}/q \\
\vdots &&& qt & rt & y_2 & y_3 & \cdots & y_{(n-1)/2} & -sy_{(n-3)/2}/q \\
\vdots && qt & x_2 & y_2 & 0 &&&& \vdots \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & x_3 & y_3 &&&&& \vdots \\
\vdots && \vdots & \vdots & \vdots &&&&& \vdots \\
0 & \cdots & 0 & x_{(n-1)/2} & y_{(n-1)/2} &&&&& \vdots \\
0 & \cdots & 0 & -sx_{(n-3)/2}/q & -sy_{(n-3)/2}/q & \cdots& \cdots & \cdots & \cdots & 0
\end{array}
\right].
$}
\]

Then, since $W(t)_{n_1+1}={}^tS(t)_{n_1+1}W(t)_{n_1}S(t)_{n_1+1}$, where

\begin{align*}
S(t)_{n_1+1}=\prod_{m=(n+3)/2}^{n}R_n\left(\frac{n+1}{2},m;-\dfrac{w_{(n-1)/2, m}(t)_{n_1}}{qt}\right)
\end{align*}
and
$w_{n-1, (n+1)/2}(t)_{n_1+1}=w_{(n+1)/2, n-1}(t)_{n_1+1}=x_{(n+1)/2}$,
we have
\begin{small}
\begin{align*}
w_{n-1, n-1}(t)_{n_1+1} & =-\frac{x_{(n-1)/2}}{qt}\cdot y_{(n-1)/2}-\frac{x_{(n-1)/2}}{qt}\cdot x_{(n+1)/2} \\[1mm]
&=-\frac{x_{(n-1)/2}}{qt}\cdot \left(-\frac{s}{q}x_{(n-3)/2} \right)-\frac{x_{(n-1)/2}}{qt}\cdot x_{(n+1)/2} \\[1mm]
&=-\frac{1}{qt} \cdot \frac{(-1)^{(n-1)/2}\left\{ \left( r+\sqrt{r^2-4qs} \right)^{(n-1)/2}-\left( r-\sqrt{r^2-4qs} \right)^{(n-1)/2} \right\}}{2^{(n-1)/2}q^{(n-5)/2}\sqrt{r^2-4qs}}t \\[2mm]
& \hspace{7mm} \times \left( -\frac{s}{q} \right)\cdot \frac{(-1)^{(n-3)/2}\left\{ \left( r+\sqrt{r^2-4qs} \right)^{(n-3)/2}-\left( r-\sqrt{r^2-4qs} \right)^{(n-3)/2} \right\}}{2^{(n-3)/2}q^{(n-7)/2}\sqrt{r^2-4qs}}t \\[2mm]
& \hspace{3mm} -\frac{1}{qt} \cdot \frac{(-1)^{(n-1)/2}\left\{ \left( r+\sqrt{r^2-4qs} \right)^{(n-1)/2}-\left( r-\sqrt{r^2-4qs} \right)^{(n-1)/2} \right\}}{2^{(n-1)/2}q^{(n-5)/2}\sqrt{r^2-4qs}}t \\[2mm]
& \hspace{7mm} \times \frac{(-1)^{(n+1)/2}\left\{ \left( r+\sqrt{r^2-4qs} \right)^{(n+1)/2}-\left( r-\sqrt{r^2-4qs} \right)^{(n+1)/2} \right\}}{2^{(n+1)/2}q^{(n-3)/2}\sqrt{r^2-4qs}}t \\[2mm]
&=-s\biggl\{\left( r+\sqrt{r^2-4qs} \right)^{n-2}-\left(4qs\right)^{(n-3)/2}\left(r+\sqrt{r^2-4qs}\right) \\[1mm]
&\hspace{7mm}-\left(4qs\right)^{(n-3)/2}\left(r-\sqrt{r^2-4qs}\right)+\left( r-\sqrt{r^2-4qs} \right)^{n-2}\biggr\}t \biggl/\left\{2^{n-2}q^{n-4}(r^2-4qs)\right\} \\[1mm]
&\hspace{3.5mm}+\biggl\{\left( r+\sqrt{r^2-4qs} \right)^{n}-\left(4qs\right)^{(n-1)/2}\left(r+\sqrt{r^2-4qs}\right) \\[1mm]
&\hspace{7mm}-\left(4qs\right)^{(n-1)/2}\left(r-\sqrt{r^2-4qs}\right)+\left( r-\sqrt{r^2-4qs} \right)^{n}\biggr\}t \biggl/\left\{2^{n}q^{n-3}(r^2-4qs)\right\} \\[1mm]
\end{align*}

\begin{align*}
&=\Biggl\{\left( r+\sqrt{r^2-4qs} \right)^{n-2}\left\{\left( r+\sqrt{r^2-4qs} \right)^{2}-4qs\right\} \\[1mm]
&\hspace{7mm}+\left( r-\sqrt{r^2-4qs} \right)^{n-2}\left\{\left( r-\sqrt{r^2-4qs} \right)^{2}-4qs\right\}\Biggr\}t\biggl/\left\{2^{n}q^{n-3}(r^2-4qs)\right\} \\[2mm]
&=\frac{2\sqrt{r^2-4qs}\left\{ \left( r+\sqrt{r^2-4qs} \right)^{n-1}-\left( r-\sqrt{r^2-4qs} \right)^{n-1} \right\}}{2^{n}q^{n-3}(r^2-4qs)}t \\[2mm]
&=\frac{\left\{ \left( r+\sqrt{r^2-4qs} \right)^{n-1}-\left( r-\sqrt{r^2-4qs} \right)^{n-1} \right\}}{2^{n-1}q^{n-3}\sqrt{r^2-4qs}}t =x_{n-1}.
\end{align*}
\end{small}

Similarly, we have

\begin{align*}
w_{n-1, n}(t)_{n_1+1} &=w_{n, n-1}(t)_{n_1+1}=-\frac{x_{(n-1)/2}}{qt}\cdot\frac{-sy_{(n-3)/2}}{q}-\frac{-sx_{(n-3)/2}}{q^2t} x_{(n+1)/2} \\[1mm]
&=-\frac{s^2}{q^3t}\cdot\frac{(-1)^{(n-1)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n-1)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n-1)/2} \right\}}{2^{(n-1)/2}q^{(n-5)/2}\sqrt{r^2-4qs}}t \\[1mm]
&\hspace{10mm} \times \frac{(-1)^{(n-5)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n-5)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n-5)/2}  \right\}}{2^{(n-5)/2}q^{(n-9)/2}\sqrt{r^2-4qs}}t \\[1mm]
&\hspace{3mm} +\frac{s}{q^2t}\cdot \frac{(-1)^{(n-3)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n-3)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n-3)/2} \right\}}{2^{(n-3)/2}q^{(n-7)/2}\sqrt{r^2-4qs}}t \\[1mm]
&\hspace{10mm} \times \frac{(-1)^{(n+1)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n+1)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n+1)/2}  \right\}}{2^{(n+1)/2}q^{(n-3)/2}\sqrt{r^2-4qs}}t \\[1mm]
&=-\frac{s^2}{q}\cdot\biggl\{(r+\sqrt{r^2-4qs})^{n-3}-(4qs)^{(n-5)/2}(r+\sqrt{r^2-4qs})^2 \\
&\hspace{7mm}-(4qs)^{(n-5)/2}(r-\sqrt{r^2-4qs})^2+(r-\sqrt{r^2-4qs})^{n-3}\biggr\}t\biggl/\left\{2^{n-3}q^{n-5}(r^2-4qs)\right\} \\
&\hspace{3mm}+\frac{s}{q}\cdot\biggl\{(r+\sqrt{r^2-4qs})^{n-1}-(4qs)^{(n-3)/2}(r+\sqrt{r^2-4qs})^2 \\
&\hspace{7mm}-(4qs)^{(n-3)/2}(r-\sqrt{r^2-4qs})^2+(r-\sqrt{r^2-4qs})^{n-1}\biggr\}t\biggl/\left\{2^{n-1}q^{n-4}(r^2-4qs)\right\} \\
&=\frac{s}{q}\cdot\frac{(r+\sqrt{r^2-4qs})^{n-2}-(r-\sqrt{r^2-4qs})^{n-2}}{2^{n-2}q^{n-4}\sqrt{r^2-4qs}}t \\
&=-\frac{s}{q}x_{n-2}=y_{n-1}.
\end{align*}

Moreover, since $w_{n, (n+1)/2}(t)_{n_1+1}=w_{(n+1)/2,n}(t)_{n_1+1}=-sx_{(n-1)/2}/q$, we have

\begin{align*}
w_{n,n}(t)_{n_1+1} &=-\frac{-sx_{(n-3)/2}}{q^2t}\cdot\frac{-sy_{(n-3)/2}}{q}-\frac{-sx_{(n-3)/2}}{q^2t}\cdot\frac{-sx_{(n-1)/2}}{q} \\[1mm]
&=\frac{s^3}{q^4t}\cdot\frac{(-1)^{(n-3)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n-3)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n-3)/2} \right\}}{2^{(n-3)/2}q^{(n-7)/2}\sqrt{r^2-4qs}}t \\[1mm]
&\hspace{21mm} \times \frac{(-1)^{(n-5)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n-5)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n-5)/2}  \right\}}{2^{(n-5)/2}q^{(n-9)/2}\sqrt{r^2-4qs}}t \\[1mm]
&\hspace{3mm} -\frac{s^2}{q^3t}\cdot \frac{(-1)^{(n-3)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n-3)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n-3)/2} \right\}}{2^{(n-3)/2}q^{(n-7)/2}\sqrt{r^2-4qs}}t \\[1mm]
&\hspace{21mm} \times \frac{(-1)^{(n-1)/2}\left\{ \left(r+\sqrt{r^2-4qs}\right)^{(n-1)/2}-\left(r-\sqrt{r^2-4qs}\right)^{(n-1)/2}  \right\}}{2^{(n-1)/2}q^{(n-5)/2}\sqrt{r^2-4qs}}t \\
& =-\frac{s^3}{q^2}\cdot\biggl\{(r+\sqrt{r^2-4sq})^{n-4}-(4qs)^{(n-5)/2}(r+\sqrt{r^2-4qs}) \\
&\hspace{7mm}-(4qs)^{(n-5)/2}(r-\sqrt{r^2-4qs})+(r-\sqrt{r^2-4qs})^{n-4}\biggr\}t\biggl/\left\{2^{n-4}q^{n-6}(r^2-4qs)\right\} \\
&\hspace{3mm}+\frac{s^2}{q^2}\cdot\biggl\{(r+\sqrt{r^2-4sq})^{n-2}-(4qs)^{(n-3)/2}(r+\sqrt{r^2-4qs}) \\
&\hspace{7mm}-(4qs)^{(n-3)/2}(r-\sqrt{r^2-4qs})+(r-\sqrt{r^2-4qs})^{n-2}\biggr\}t\biggl/\left\{2^{n-2}q^{n-5}(r^2-4qs)\right\} \\
&=\frac{s^2}{q^2}\cdot\frac{(r+\sqrt{r^2-4sq})^{n-3}-(r-\sqrt{r^2-4qs})^{n-3}}{2^{n-3}q^{n-5}\sqrt{r^2-4qs}}t \\[2mm]
&=-\frac{s}{q}\left(-\frac{s}{q}x_{n-3}\right)=-\frac{s}{q}y_{n-2}.
\end{align*}
\normalsize

Therefore, we can express the matrix $W(t)_{n_0}$ $(=W(t)_{n_1+1})$ as follows;
\begin{align*}
W(t)_{n_0}=
\scalebox{1}[1]
{$\left[
\begin{array}{ccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&&& \rotatebox[origin=c]{300}{\vdots} && \vdots & \vdots \\
\vdots &&&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&&& qt & 0 & 0 & \cdots & 0 & 0 \\
\vdots &&& qt & * & * & * & \cdots & * & * \\
\vdots && qt & 0 & * & * & * & \cdots & * & * \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & 0 & * & * & * & \cdots & * & * \\
\vdots && \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & \cdots & 0 & 0 & * & * & * & \cdots & x_{n-1} & y_{n-1} \\
0 & \cdots & 0 & 0 & * & * & * & \cdots & y_{n-1} & (-s/q)y_{n-2}
\end{array}
\right].
$}
\end{align*}
Then, by the definition of the matrix $W_{k}$ $(n_0< k \leq n-2)$, we have
\begin{align*}
W(t)_{n-2}=
\scalebox{1}[1]
{$\left[
\begin{array}{cccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&& \rotatebox[origin=c]{310}{\vdots} && \vdots & \vdots \\
\vdots &&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&& qt & 0 & 0 & \cdots & 0 & 0 \\
\vdots && qt & 0 & 0 & 0 & \cdots & 0 & 0 \\
\vdots & \rotatebox[origin=c]{310}{\vdots} & 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\
\vdots && \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 && 0 & 0 & 0 & 0 & \cdots & x_{n-1} & y_{n-1} \\[1.5mm]
0 & \cdots & 0 & 0 & 0 & 0 & \cdots & y_{n-1} & (-s/q)y_{n-2} \\
\end{array}
\right],
$}
\end{align*}
which completes the proof of  \cref{lem6.4} for odd $n$.

Next, suppose $n$ is even. Then, we have $n_1=(n-4)/2$ and hence by  \cref{lem6.2}, we have
\begin{align*}
w_{ij}(t)_{n_1}
=
\begin{cases}
0 & \text{$(i,j)=((n-4)/2, \ell)$ or $(\ell, (n-4)/2)$ $((n+6)/2\leq \ell \leq n)$}, \\
x_{\ell-n/2} & \text{$(i,j)=((n-2)/2, \ell)$ or $(\ell, (n-2)/2)$ $((n+4)/2 \leq \ell \leq n-1)$}, \\
y_{\ell-n/2} & \text{$(i,j)=(n/2, \ell)$ or $(\ell, n/2)$ $((n+4)/2 \leq \ell \leq n-1)$}, \\
-sx_{(n-4)/2}/q & \text{$(i,j)=((n-2)/2, n)$ or $(n, (n-2)/2)$}, \\
-sy_{(n-4)/2}/q & \text{$(i,j)=(n/2, n)$ or $(n, n/2)$}, \\
w_{ij}(t)_{(n-6)/2} & \text{otherwise}.
\end{cases}
\end{align*}
\normalsize

Thus, we have the next expression of the matrix $W(t)_{n_1}$;

\begin{align*}
W(t)_{n_1}
=
\scalebox{0.8}[1]
{$\left[
\begin{array}{cccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&&&& \rotatebox[origin=c]{300}{\vdots} && \vdots & \vdots \\
\vdots &&&&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&&&& qt & x_2 & x_3 & \cdots & x_{(n-2)/2} & -sx_{(n-4)/2}/q \\
\vdots &&&& qt & rt & y_2 & y_3 & \cdots & y_{(n-2)/2} & -sy_{(n-4)/2}/q \\
\vdots &&& qt & rt & st & 0 & 0 & \cdots & 0 & 0 \\
\vdots && qt & x_2 & y_2 & 0 & 0 &&&& \vdots \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & x_3 & y_3 & 0 &&&&& \vdots \\
\vdots && \vdots & \vdots & \vdots & \vdots &&&&& \vdots \\
0 & \cdots & 0 & x_{(n-2)/2} & y_{(n-2)/2} & 0 &&&&& \vdots \\
0 & \cdots & 0 & -sx_{(n-4)/2}/q & -sy_{(n-4)/2}/q & 0 & \cdots& \cdots & \cdots & \cdots & 0
\end{array}
\right].
$}
\end{align*}

Then, for any integer $\ell$ $((n+4)/2 \leq \ell \leq n)$, let us put
\begin{align*}
V(t)_{\ell}=(v_{ij}(t)_{\ell})_{1\leq i,j \leq n}={}^tR(t)_{\ell}W(t)_{n_1}R(t)_{\ell},
\end{align*}
where
\begin{align*}
R(t)_{\ell}=\prod_{m=(n+4)/2}^{\ell}R_n\left((n+2)/2,m;-\dfrac{w_{(n-2)/2, m}(t)_{n_1}}{qt}\right).
\end{align*}
Note that we have $V(t)_n=W(t)_{n_1+1}$.

\begin{clm}
For any integer $\ell^{\prime}$ $((n+4)/2 \leq \ell^{\prime} \leq n-1)$, we have

\begin{align*}
v_{ij}(t)_{\ell^{\prime}}
=
\begin{cases}
0 & \text{$(i,j)=((n-2)/2,\ell)$ or $(\ell,(n-2)/2)$ $((n+4)/2 \leq \ell \leq \ell^{\prime})$}, \\
x_{\ell-(n-2)/2} & \text{$(i,j)=(n/2, \ell)$ or $(\ell, n/2)$ $((n+2)/2 \leq \ell \leq \ell^{\prime})$}, \\
y_{\ell-(n-2)/2} & \text{$(i,j)=((n+2)/2, \ell)$ or $(\ell, (n+2)/2)$ $((n+2)/2 \leq \ell \leq \ell^{\prime})$}, \\
\left( -x_{\ell-n/2}/qt \right)^2y_2 & \text{$(i,j)=(\ell, \ell)$ $((n+4)/2 \leq \ell \leq \ell^{\prime})$}, \\
(-x_{m-n/2}/qt) \\
\hspace{6.5mm}\times y_{\ell-(n-2)/2} & \text{$(i,j)=(\ell,m)$ or $(m, \ell)$ $((n+4)/2 \leq \ell<m \leq \ell^{\prime})$}, \\
w_{ij}(t)_{(n-4)/2} & \text{otherwise}.
\end{cases}
\end{align*}
\normalsize
\end{clm}

\begin{proof}
To ease notation, let us put $k=(n-2)/2$. Then, by definition,
\normalsize
\begin{align*}
V(t)_{(n+4)/2}=
\scalebox{0.75}[0.95]
{$\left[
\begin{array}{cccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&&&& \rotatebox[origin=c]{300}{\vdots} && \vdots & \vdots \\
\vdots &&&&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&&&& qt & 0 & x_3 & \cdots & x_{k} & -sx_{k-1}/q \\
\vdots &&&& qt & x_2 & x_3 & y_3 & \cdots & y_{k} & -sy_{k-1}/q \\
\vdots &&& qt & x_2 & y_2 & y_3 & 0 & \cdots & 0 & 0 \\
\vdots && qt & 0 & x_3 & y_3 & \left( -x_2/qt \right)^2y_2 &&&& \vdots \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & x_3 & y_3 & 0 &&&&& \vdots \\
\vdots & & \vdots & \vdots & \vdots & \vdots &&&&& \vdots \\
0 & \cdots & 0 & x_{k} & y_{k} & 0 &&&&& \vdots \\
0 & \cdots & 0 & -sx_{k-1}/q & -sy_{k-1}/q & 0 & \cdots& \cdots & \cdots & \cdots & 0
\end{array}
\right]
$}
\end{align*}
\normalsize
and we get Claim for $V(t)_{(n+4)/2}$. Here, suppose Claim is true for $V(t)_{\ell^{\prime}-1}$ and hence

\begin{align*}
V(t)_{\ell^{\prime}-1}=
\scalebox{0.63}[0.8]
{$\left[
\begin{array}{cccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 & \cdots \\
\vdots &&&&& \rotatebox[origin=c]{300}{\vdots} &&& \vdots & \vdots & \\
\vdots &&&& qt & 0 & 0 & \cdots & 0 & x_{\ell^{\prime}-k-1} & \cdots \\
\vdots &&& qt & x_2 & x_3 & x_4 & \cdots & x_{\ell^{\prime}-k-1} & y_{\ell^{\prime}-k-1} & \cdots \\
\vdots && qt & x_2 & y_2 & y_3 & y_4 & \cdots & y_{\ell^{\prime}-k-1} & 0 & \cdots \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & x_3 & y_3 & \left( -x_2/qt \right)^2y_2 & \left( -x_3/qt \right)y_3 & \cdots & (-x_{\ell^{\prime}-k-2}/qt)y_3 & 0 & \cdots \\
\vdots && 0 & x_4 & y_4 & \left( -x_3/qt \right)y_3 & \left( -x_3/qt \right)^2y_2 & \cdots & (-x_{\ell^{\prime}-k-2}/qt)y_4 & 0 & \cdots \\
\vdots && \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \\
0 && 0 & x_{\ell^{\prime}-k-1} & y_{\ell^{\prime}-k-1} & (-x_{\ell^{\prime}-k-2}/qt)y_3 & (-x_{\ell^{\prime}-k-2}/qt)y_4 & \cdots & (-x_{\ell^{\prime}-k-2}/qt)^2y_2 & 0 & \cdots \\
0 & \cdots & x_{\ell^{\prime}-k-1} & y_{\ell^{\prime}-k-1} & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots \\
\vdots && \vdots & \vdots & \vdots & \vdots & \vdots && \vdots & \vdots
\end{array}
\right].
$}
\end{align*}

Thus, by a direct computation, we have

\begin{align*}
V(t)_{\ell^{\prime}}=
\scalebox{0.7}[0.9]
{$\left[
\begin{array}{cccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 & \cdots \\
\vdots &&&&& \rotatebox[origin=c]{300}{\vdots} &&& \vdots & \vdots & \\
\vdots &&&& qt & 0 & 0 & \cdots & 0 & x_{\ell^{\prime}-k} & \cdots \\
\vdots &&& qt & x_2 & x_3 & x_4 & \cdots & x_{\ell^{\prime}-k} & y_{\ell^{\prime}-k} & \cdots \\
\vdots && qt & x_2 & y_2 & y_3 & y_4 & \cdots & y_{\ell^{\prime}-k} & 0 & \cdots \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & x_3 & y_3 & \left( -x_2/qt \right)^2y_2 & \left( -x_3/qt \right)y_3 & \cdots & (-x_{\ell^{\prime}-k-1}/qt)y_3 & 0 & \cdots \\
\vdots && 0 & x_4 & y_4 & \left( -x_3/qt \right)y_3 & \left( -x_3/qt \right)^2y_2 & \cdots & (-x_{\ell^{\prime}-k-1}/qt)y_4 & 0 & \cdots \\
\vdots && \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \\
0 && 0 & x_{\ell^{\prime}-k} & y_{\ell^{\prime}-k} & (-x_{\ell^{\prime}-k-1}/qt)y_3 & (-x_{\ell^{\prime}-k-1}/qt)y_4 & \cdots & (-x_{\ell^{\prime}-k-1}/qt)^2y_2 & 0 & \cdots \\
0 & \cdots & x_{\ell^{\prime}-k} & y_{\ell^{\prime}-k} & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots \\
\vdots && \vdots & \vdots & \vdots & \vdots & \vdots && \vdots & \vdots
\end{array}
\right]
$}
\end{align*}
\normalsize
and we get Claim by induction on $\ell^{\prime}$.
\end{proof}

By above Claim,

\begin{align*}
V(t)_{n-1}=
\scalebox{0.7}[0.7]
{$\left[
\begin{array}{cccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&& \rotatebox[origin=c]{300}{\vdots} &&& \vdots & \vdots \\
\vdots &&&& qt & 0 & 0 & \cdots & 0 & -sx_{k-1}/q \\
\vdots &&& qt & x_2 & x_3 & x_4 & \cdots & x_{k+1} & -sy_{k-1}/q \\
\vdots && qt & x_2 & y_2 & y_3 & y_4 & \cdots & y_{k+1} & 0 \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & x_3 & y_3 & \left( -x_2/qt \right)^2y_2 & \left( -x_3/qt \right)y_3 & \cdots & (-x_{k}/qt)y_3 & 0 \\
\vdots && 0 & x_4 & y_4 & \left( -x_3/qt \right)y_3 & \left( -x_3/qt \right)^2y_2 & \cdots & (-x_{k}/qt)y_4 & 0 \\
\vdots && \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 && 0 & x_{k+1} & y_{k+1} & (-x_{k}/qt)y_3 & (-x_{k}/qt)y_4 & \cdots & (-x_{k}/qt)^2y_2 & 0 \\
0 & \cdots & -sx_{k-1}/q & -sy_{k-1}/q & 0 & 0 & 0 & \cdots & 0 & 0 \\
\end{array}
\right]
$}
\end{align*}
\normalsize
and hence, by the definition of $V(t)_n$, we have

\begin{align*}
&W(t)_{n_1+1}=V(t)_{n}= \\
&\scalebox{0.7}[0.7]
{$\left[
\begin{array}{cccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&& \rotatebox[origin=c]{300}{\vdots} &&& \vdots & \vdots \\
\vdots &&&& qt & 0 & 0 & \cdots & 0 & 0 \\
\vdots &&& qt & x_2 & x_3 & x_4 & \cdots & x_{k+1} & -sx_{k}/q \\
\vdots && qt & x_2 & y_2 & y_3 & y_4 & \cdots & y_{k+1} & -sy_{k}/q \\
\vdots & \rotatebox[origin=c]{300}{\vdots} & 0 & x_3 & y_3 & \left( -x_2/qt \right)^2y_2 & \left( -x_3/qt \right)y_3 & \cdots & (-x_{k}/qt)y_3 & (x_{k-1}/q^2t^2)y_2y_3 \\
\vdots && 0 & x_4 & y_4 & \left( -x_3/qt \right)y_3 & \left( -x_3/qt \right)^2y_2 & \cdots & (-x_{k}/qt)y_4 & (x_{k-1}/q^2t^2)y_2y_4 \\
\vdots && \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 && 0 & x_{k+1} & y_{k+1} & (-x_{k}/qt)y_3 & (-x_{k}/qt)y_4 & \cdots & (-x_{k}/qt)^2y_2 & (x_{k-1}/q^2t^2)y_2y_{k+1} \\[1.5mm]
0 & \cdots & 0 & -sx_{k}/q & -sy_{k}/q & (x_{k-1}/q^2t^2)y_2y_3 & (x_{k-1}/q^2t^2)y_2y_4 & \cdots & (x_{k-1}/q^2t^2)y_2y_{k+1} & (x_{k-1}/q^2t^2)^2y_2^3 \\
\end{array}
\right].
$}
\end{align*}

Therefore, by the definition of the matrix $W(t)_{n_1+2}=W(t)_{n_0}$, we have
\begin{small}
\begin{align*}
w_{n-1,n-1}(t)_{n_0} &=\left(-\frac{x_{(n-2)/2}}{qt}\right)^2y_2-\frac{\left(x_{n/2}\right)^2}{qt} \\[2mm]
&=\frac{s\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-2}+\left(r-\sqrt{r^2-4qs}\right)^{n-2}-2(4qs)^{(n-2)/2}\right\}}{2^{n-2}q^{n-4}\left(r^2-4qs\right)}t \\
&\hspace{45mm}-\frac{\left(r+\sqrt{r^2-4qs}\right)^{n}+\left(r-\sqrt{r^2-4qs}\right)^{n}-2(4qs)^{n/2}}{2^nq^{n-3}\left(r^2-4qs\right)}t \\
&=\frac{\left(r+\sqrt{r^2-4qs}\right)^{n-2}\left\{ 4qs-\left(r+\sqrt{r^2-4qs}\right)^2\right\}}{2^nq^{n-3}\left(r^2-4qs\right)}t \\
&\hspace{45mm}+\frac{\left(r-\sqrt{r^2-4qs}\right)^{n-2}\left\{ 4qs-\left(r-\sqrt{r^2-4qs}\right)^2 \right\}}{2^nq^{n-3}\left(r^2-4qs\right)}t \\
&=\frac{\left(r+\sqrt{r^2-4qs}\right)^{n-2}\left(-2\sqrt{r^2-4qs}\right)\left(r+\sqrt{r^2-4qs}\right)}{2^nq^{n-3}\left(r^2-4qs\right)}t \\
&\hspace{40mm}+\frac{\left(r-\sqrt{r^2-4qs}\right)^{n-2}\left(2\sqrt{r^2-4qs}\right)\left(r-\sqrt{r^2-4qs}\right)}{2^nq^{n-3}\left(r^2-4qs\right)}t =x_{n-1}.
\end{align*}
\end{small}

Similarly, we have

\begin{align*}
&w_{n-1,n}(t)_{n_0}=w_{n,n-1}(t)_{n_0}=\frac{x_{(n-4)/2}}{q^2t^2}y_2y_{n/2}+\frac{sx_{(n-2)/2}x_{n/2}}{q^2t} \\
&=\frac{x_{(n-4)/2}}{q^2t^2}y_2\left(-\frac{s}{q}x_{(n-2)/2}\right)+\frac{sx_{(n-2)/2}x_{n/2}}{q^2t} \\
&=-\frac{s^2}{q^3}\cdot\frac{-\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-3}+\left(r-\sqrt{r^2-4qs}\right)^{n-3}-2r\left( 4qs \right)^{(n-4)/2}\right\}}{2^{n-3}q^{n-7}\left(r^2-4qs\right)}t \\
&\hspace{17mm}+\frac{s}{q^2}\cdot\frac{-\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-1}+\left(r-\sqrt{r^2-4qs}\right)^{n-1}-2r\left( 4qs \right)^{(n-2)/2}\right\}}{2^{n-1}q^{n-5}\left(r^2-4qs\right)}t \\
&=\frac{s\left\{4qs\left(r+\sqrt{r^2-4qs}\right)^{n-3}+4qs\left(r-\sqrt{r^2-4qs}\right)^{n-3}-2r\left( 4qs \right)^{(n-2)/2}\right\}}{2^{n-1}q^{n-3}\left(r^2-4qs\right)}t \\
&\hspace{22mm}-\frac{s\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-1}+\left(r-\sqrt{r^2-4qs}\right)^{n-1}-2r\left( 4qs \right)^{(n-2)/2}\right\}}{2^{n-1}q^{n-3}\left(r^2-4qs\right)}t \\
&=\frac{s\left(r+\sqrt{r^2-4qs}\right)^{n-3}\left(-2\sqrt{r^2-4qs}\right)\left(r+\sqrt{r^2-4qs}\right)}{2^{n-1}q^{n-3}\left(r^2-4qs\right)}t \\
&\hspace{25mm}+\frac{s\left(r-\sqrt{r^2-4qs}\right)^{n-3}\left(2\sqrt{r^2-4qs}\right)\left(r-\sqrt{r^2-4qs}\right)}{2^{n-1}q^{n-3}\left(r^2-4qs\right)}t \\
&=-\frac{s}{q}\cdot\frac{\left(r+\sqrt{r^2-4qs}\right)^{n-2}-\left(r-\sqrt{r^2-4qs}\right)^{n-2}}{2^{n-2}q^{n-4}\sqrt{r^2-4qs}}t=y_{n-1},
\end{align*}

\begin{small}
\begin{align*}
w_{n,n}(t)_{n_0} &=\left(\frac{x_{(n-4)/2}}{q^2t^2}\right)^2y_2^3-\frac{s^2\left(x_{(n-2)/2}\right)^2}{q^3t} \\
&=\frac{s^3\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-4}+\left(r-\sqrt{r^2-4qs}\right)^{n-4}-2(4qs)^{(n-4)/2}\right\}}{2^{n-4}q^{n-4}(r^2-4qs)}t \\
&\hspace{30mm}-\frac{s^2\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-2}+\left(r-\sqrt{r^2-4qs}\right)^{n-2}-2(4qs)^{(n-2)/2}\right\}}{2^{n-2}q^{n-3}(r^2-4qs)}t \\
&=\frac{s^2\left\{4qs\left(r+\sqrt{r^2-4qs}\right)^{n-4}+4qs\left(r-\sqrt{r^2-4qs}\right)^{n-4}-2(4qs)^{(n-2)/2}\right\}}{2^{n-2}q^{n-3}(r^2-4qs)}t \\
&\hspace{30mm}-\frac{s^2\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-2}+\left(r-\sqrt{r^2-4qs}\right)^{n-2}-2(4qs)^{(n-2)/2}\right\}}{2^{n-2}q^{n-3}(r^2-4qs)}t \\
&=\frac{s^2\left(r+\sqrt{r^2-4qs}\right)^{n-4}\left(-2\sqrt{r^2-4qs}\right)\left(r+\sqrt{r^2-4qs}\right)}{2^{n-2}q^{n-3}\left(r^2-4qs\right)}t \\
&\hspace{25mm}+\frac{s^2\left(r-\sqrt{r^2-4qs}\right)^{n-4}\left(2\sqrt{r^2-4qs}\right)\left(r-\sqrt{r^2-4qs}\right)}{2^{n-2}q^{n-3}\left(r^2-4qs\right)}t \\
&=\left(-\frac{s}{q} \right)\cdot\left(-\frac{s}{q} \right) \cdot \frac{-\left\{\left(r+\sqrt{r^2-4qs}\right)^{n-3}-\left(r-\sqrt{r^2-4qs}\right)^{n-3}\right\}}{2^{n-3}q^{n-5}\sqrt{r^2-4qs}}t =-\frac{s}{q}y_{n-2}.
\end{align*}

\end{small}

Therefore, we can express the matrix $W(t)_{n_0}$ as follows;
\begin{align*}
W(t)_{n_0}=
\scalebox{0.75}[0.75]
{$\left[
\begin{array}{cccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&& \rotatebox[origin=c]{305}{\vdots} && \vdots & \vdots \\
\vdots &&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&& qt & 0 & 0 & \cdots & 0 & 0 \\
\vdots && qt & 0 & * & * & \cdots & * & * \\
\vdots & \rotatebox[origin=c]{305}{\vdots} & 0 & 0 & * & * & \cdots & * & * \\
\vdots && \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 && 0 & 0 & * & * & \cdots & x_{n-1} & y_{n-1} \\[1.5mm]
0 & \cdots & 0 & 0 & * & * & \cdots & y_{n-1} & (-s/q)y_{n-2} \\
\end{array}
\right].
$}
\end{align*}
Then, by the definition of the matrix $W_{k}$ $(n_0 < k \leq n-2)$, we have
\begin{align*}
W(t)_{n-2}=
\scalebox{0.75}[0.75]
{$\left[
\begin{array}{cccccccccccccc}
0 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 0 \\
\vdots &&&&& \rotatebox[origin=c]{310}{\vdots} && \vdots & \vdots \\
\vdots &&&& qt & 0 & \cdots & 0 & 0 \\
\vdots &&& qt & 0 & 0 & \cdots & 0 & 0 \\
\vdots && qt & 0 & 0 & 0 & \cdots & 0 & 0 \\
\vdots & \rotatebox[origin=c]{310}{\vdots} & 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\
\vdots && \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 && 0 & 0 & 0 & 0 & \cdots & x_{n-1} & y_{n-1} \\[1.5mm]
0 & \cdots & 0 & 0 & 0 & 0 & \cdots & y_{n-1} & (-s/q)y_{n-2} \\
\end{array}
\right],
$}
\end{align*}
which completes the proof of  \cref{lem6.4} for even $n$.
\end{proof}
In the end, let us apply  \cref{lem6.4} to the matrix $A_{\bm c}(t)_1$ $(n \geq 5)$. Then, by the definition of the matrix $A_{\bm c}(t)_k$ $(2 \leq k \leq n-2)$ and  \cref{lem6.4}, we have

\begin{align} \label{eq6.4}
A_{\bm c}(t)_{n-2}
=
\scalebox{0.75}[0.75]
{$\left[
\begin{array}{cccccccc}
1 & 0 & \dots & \dots & \dots & 0 & 0 & 0 \\
0 &&&&& -(n-2)t & 0 & 0 \\
\vdots &&&&  \rotatebox[origin=c]{292}{\vdots} & \rotatebox[origin=c]{292}{\vdots} & 0 & 0 \\
\vdots &&& \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \rotatebox[origin=c]{292}{\vdots} & \vdots & \vdots \\
\vdots && \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} & \vdots & \vdots \\
0 & -(n-2)t & \rotatebox[origin=c]{292}{\vdots} & \rotatebox[origin=c]{292}{\vdots} & \hspace{3mm}  \rotatebox[origin=c]{292}{\vdots} && 0 & 0 \\
0 & 0 & 0 & \dots & \dots & 0 & a_{n-1,n-1}^{({\bm c})}(t)_{n-2} & a_{n-1,n}^{({\bm c})}(t)_{n-2} \\
0 & 0 & 0 & \dots & \dots & 0 & a_{n,n-1}^{({\bm c})}(t)_{n-2} & a_{n,n}^{({\bm c})}(t)_{n-2}
\end{array}
\hspace{-0.5mm}\right],
$}
\end{align}
\normalsize
where
\begin{align*}
\begin{cases}
a_{n-1,n-1}^{(\bm c)}(t)_{n-2}=2\left(1-2/n \right)t^2+\bar{x}_{n-1} \\[2mm]
a_{n-1,n}^{(\bm c)}(t)_{n-2}=a_{n,n-1}^{(\bm c)}(t)_{n-2}=\left(1-2/n\right)at^2-(nb\bar{x}_{n-2})/(n-2) \\[2mm]
a_{n,n}^{(\bm c)}(t)_{n-2}=\left\{\left(1-1/n\right)a^2-2b\right\}t^2+(n^2b^2\bar{x}_{n-3})/(n-2)^2.
\end{cases}
\end{align*}
Here, for any integer $m \geq 0$, we denote
\begin{align*}
&\bar{x}_{m}
=
\frac{(-1)^m\left(A^m-B^m\right)}{2^m\{-(n-2)\}^{m-2}\sqrt{(n-1)^2a^2-4n(n-2)b}}t, \\[2mm]
&A=-(n-1)a+\sqrt{(n-1)^2a^2-4n(n-2)b}, \\[2mm]
&B=-(n-1)a-\sqrt{(n-1)^2a^2-4n(n-2)b}.
\end{align*}

\section{Proof of Theorems}
By a direct computation, we have

\begin{align} \label{eq6.5}
&\det
\left[
\begin{array}{cc}
a_{n-1,n-1}^{({\bm c})}(t)_{n-2} & a_{n-1,n}^{({\bm c})}(t)_{n-2} \\ \notag
a_{n,n-1}^{({\bm c})}(t)_{n-2} & a_{n,n}^{({\bm c})}(t)_{n-2}
\end{array}
\right] \\[1mm] \notag
&=\left(1-\frac{2}{n}\right)(a^2-4b)t^4 \\[1mm] \notag
&\hspace{2.5mm}+\frac{2(n-2)}{n}t^2\cdot\frac{n^2b^2\bar{x}_{n-3}}{(n-2)^2}+\left\{ \left( \frac{n-1}{n}a^2-2b \right) \right\}t^2\cdot\bar{x}_{n-1}+\frac{2(n-2)}{n}at^2\cdot\frac{nb\bar{x}_{n-2}}{n-2} \\[0mm] \notag
&\hspace{67.5mm}+\frac{n^2b^2}{(n-2)^2}\bar{x}_{n-1}\bar{x}_{n-3}-\frac{n^2b^2}{(n-2)^2}\bar{x}_{n-2}^2 \\[0mm]
&=\frac{(n-2)(a^2-4b)}{n}t^4+\left\{\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2}\right\}t^2 \\[0mm] \notag
&\hspace{76mm}+\frac{n^2b^2}{(n-2)^2}\left(\bar{x}_{n-1}\bar{x}_{n-3}-\bar{x}_{n-2}^2\right).
\end{align}
\normalsize
Here, let us put
\begin{align*}
&\alpha(t)=\left\{\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2}\right\}t^2, \\
&\beta(t)=\frac{n^2b^2}{(n-2)^2}\left(\bar{x}_{n-1}\bar{x}_{n-3}-\bar{x}_{n-2}^2\right).
\end{align*}
\begin{lem}\label{lem6.5}
Suppose $n\geq 5$ and put

\begin{align*}
S_k=&(n-1)^3\binom{n-k-3}{k}a^4-\frac{n(n-1)\left\{ 5n^2-(6k+23)n+10k+24 \right\}}{n-k-3}\binom{n-k-3}{k}a^2b \\
&\hspace{80mm}+4n^2(n-2)\binom{n-k-4}{k}b^2.
\end{align*}
\normalsize
Then, we have

\begin{align*}
&\alpha(t)=
\begin{cases}
\displaystyle \sum_{k=0}^{m_0}\frac{(-1)^{n+k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^kS_k}{n(n-2)^{n-3}}t^3 & \hspace{-1mm} \text{$(a \neq 0$, or $a=0$, $n : $ even$)$}, \\
0 & \hspace{-1mm} \text{$(a=0$, $n : $ odd$)$}.
\end{cases} \\
&\beta(t)=-\frac{n^{n-1}b^{n-1}}{(n-2)^{n-3}}t^2.
\end{align*}
\end{lem}
To carry out these computations, we need some combinatorial identities.
\begin{lem}{\cite[Chapter 2, Problem 18 (c), Chapter 6, Problem 18 (a)]{rio}}\label{lem6.6}
Let $N$ $(\geq 0)$ be an integer and put $\underline{m}=\lfloor N/2 \rfloor$, $\overline{m}=\lceil N/2 \rceil=\lfloor (N+1)/2 \rfloor$. Then, we have
\begin{align*}
\sum_{j=k}^{\underline{m}}\binom{N+1}{2j+1}\binom{j}{k}&=2^{N-2k}\binom{N-k}{k} \hspace{35mm} (k \in \Z, \ 0 \leq k \leq \underline{m}), \\[2mm]
\sum_{j=k}^{\overline{m}}\binom{N+1}{2j}\binom{j}{k}&=2^{N-2k}\left[2\binom{N+1-k}{k}-\binom{N-k}{k} \right] \\[2mm]
&=2^{N-2k}\left[\binom{N+1-k}{k}+\binom{N-k}{k-1} \right] \hspace{5mm} (k \in \Z, \ 0 \leq k \leq \overline{m}).
\end{align*}
\end{lem}
\begin{proof}
We omit the proof of the first identity and let us prove the second one. By using the convention
\begin{align*}
\binom{n}{n+m}=0 \hspace{3mm} (n \in \Z_{\geq 0}, \ m \in \Z_{\geq 1} ),
\end{align*}
we have
\begin{align*}
\sum_{j=k}^{\underline{m}}\binom{N+1}{2j+1}\binom{j}{k}=\sum_{j=k}^{\overline{m}}\binom{N+1}{2j+1}\binom{j}{k}
\end{align*}
and hence
\begin{align*}
\sum_{j=k}^{\overline{m}}\binom{N+1}{2j}\binom{j}{k}+\sum_{j=k}^{\underline{m}}\binom{N+1}{2j+1}\binom{j}{k}
&=\sum_{j=k}^{\overline{m}}\binom{N+1}{2j}\binom{j}{k}+\sum_{j=k}^{\overline{m}}\binom{N+1}{2j+1}\binom{j}{k} \\
&=\sum_{j=k}^{\overline{m}}\left\{ \binom{N+1}{2j}+\binom{N+1}{2j+1} \right\}\binom{j}{k} \\
&=\sum_{j=k}^{\overline{m}}\binom{N+2}{2j+1}\binom{j}{k}
\end{align*}
for any $k$ $(0 \leq k \leq \overline{m})$. Thus, by using the first identity, we get the second one.
\end{proof}

\begin{proof}[Proof of  \cref{lem6.5}]
We first prove the second equality. Then, by the definition of $\bar{x}_m$, we have
\begin{align*}
\beta(t)&=\frac{n^2b^2}{(n-2)^2}\cdot\frac{-A^{n-1}B^{n-3}-A^{n-3}B^{n-1}+2A^{n-2}B^{n-2}}{2^{2n-4}(n-2)^{2n-8}\left\{(n-1)^2a^2-4n(n-2)b\right\}}t^2 \\[2mm]
&=-\frac{n^2b^2\left\{ \left( 4n(n-2)b \right)^{n-3}\left(A^2+B^2 \right)-2\left( 4n(n-2)b \right)^{n-2} \right\}}{2^{2n-4}(n-2)^{2n-6}\left\{(n-1)^2a^2-4n(n-2)b\right\}}t^2 \\[2mm]
&=-\frac{2^{2n-6}n^{n-1}(n-2)^{n-3}b^{n-1}\cdot4\left\{(n-1)^2a^2-4n(n-2)b\right\}}{2^{2n-4}(n-2)^{2n-6}\left\{(n-1)^2a^2-4n(n-2)b\right\}}t^2 \\[2mm]
&=-\frac{n^{n-1}b^{n-1}}{(n-2)^{n-3}}t^2,
\end{align*}
which is the second equality. For the first equality, let us first suppose $a=0$. Then, we have
\begin{align*}
\bar{x}_m&=\frac{(-1)^m\left\{\left\{-4n(n-2)b\right\}^{m/2}-(-1)^m\left\{-4n(n-2)b\right\}^{m/2}\right\}}{2^m\left\{-(n-2)\right\}^{m-2}\left\{-4n(n-2)b\right\}^{1/2}}t \\
&=
\begin{cases}
\displaystyle \frac{2\left\{-4n(n-2)b\right\}^{(m-1)/2}}{2^m(n-2)^{m-2}}t & \text{($m : $ odd)}, \\
0 & \text{($m : $ even)}
\end{cases} \\
&=
\begin{cases}
\displaystyle \frac{(-nb)^{(m-1)/2}}{(n-2)^{(m-3)/2}}t & \text{($m : $ odd)}, \\
0 & \text{($m : $ even)}
\end{cases}
\end{align*}
and hence we have
\begin{align*}
\alpha(t)&=\left\{ \frac{2nb^2\bar{x}_{n-3}}{n-2}-2b\bar{x}_{n-1} \right\}t^2 \\
&=
\begin{cases}
\displaystyle \left\{ \frac{2nb^2}{n-2}\cdot\frac{(-nb)^{(n-4)/2}}{(n-2)^{(n-6)/2}}t-\frac{2b(-nb)^{(n-2)/2}}{(n-2)^{(n-4)/2}}t \right\}t^2 & \text{($n : $ even)}, \\
0 & \text{($n : $ odd)}
\end{cases} \\
&=
\begin{cases}
\displaystyle \frac{(-1)^{n/2}4n^{(n-2)/2}b^{n/2}}{(n-2)^{(n-4)/2}}t^3 & \text{($n : $ even)}, \\
0 & \text{($n : $ odd)},
\end{cases}
\end{align*}
which is the claim of   \cref{lem6.5} for the case $a=0$ since for even $n$ $(n \geq 6)$,
\begin{align*}
&\displaystyle \frac{\sum_{k=0}^{m_0}(-1)^{n+k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^kS_k}{n(n-2)^{n-3}}t^3 \\
&=\frac{(-1)^{(3n-4)/2}n^{(n-4)/2}(n-2)^{(n-4)/2}b^{(n-4)/2}\cdot4n^2(n-2)b^2}{n(n-2)^{n-3}}t^3 \\
&=\frac{(-1)^{n/2}4n^{(n-2)/2}b^{n/2}}{(n-2)^{(n-4)/2}}t^3.
\end{align*}
Next, we suppose $a\neq0$ and let us put
\begin{align*}
&F=(n-1)^2a^3-2n(2n-3)ab, \ G=(n-1)a^2-2nb, \\
&H=(n-1)^2a^2-4n(n-2)b.
\end{align*}
Then, we have
\normalsize
\begin{align*}
&\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2} \\[1mm]
&=\frac{2nb^2}{n-2}\frac{\left(A^{n-3}-B^{n-3}\right)}{2^{n-3}(n-2)^{n-5}\sqrt{H}}t+\frac{(n-1)a^2-2nb}{n}\frac{\left(A^{n-1}-B^{n-1}\right)}{2^{n-1}(n-2)^{n-3}\sqrt{H}}t \\[2mm]
&\hspace{90mm}+\frac{2ab(A^{n-2}-B^{n-2})}{2^{n-2}(n-2)^{n-4}\sqrt{H}}t \\[2mm]
&=\frac{8n^2(n-2)b^2(A^{n-3}-B^{n-3})}{2^{n-1}n(n-2)^{n-3}\sqrt{H}}t+\frac{\left\{ (n-1)a^2-2nb \right\}(A^{n-1}-B^{n-1})}{2^{n-1}n(n-2)^{n-3}\sqrt{H}}t \\[1mm]
&\hspace{78mm}+\frac{4n(n-2)ab(A^{n-2}-B^{n-2})}{2^{n-1}n(n-2)^{n-3}\sqrt{H}}t \\[1mm]
&=\frac{8n^2(n-2)b^2+\left\{ (n-1)a^2-2nb \right\}A^2+4n(n-2)abA}{2^{n-1}n(n-2)^{n-3}\sqrt{H}}A^{n-3}t \\[1mm]
&\hspace{28mm}-\frac{8n^2(n-2)b^2+\left\{ (n-1)a^2-2nb \right\}B^2+4n(n-2)abB}{2^{n-1}n(n-2)^{n-3}\sqrt{H}}B^{n-3}t
\end{align*}
\normalsize
and
\normalsize
\begin{align*}
&8n^2(n-2)b^2+\left\{ (n-1)a^2-2nb \right\}A^2+4n(n-2)abA \\
&=8n^2(n-2)b^2+\left\{ (n-1)a^2-2nb \right\}\left\{ 2(n-1)^2a^2-4n(n-2)b-2(n-1)a\sqrt{H} \right\} \\
&\hspace{70mm}+4n(n-2)ab\left\{ -(n-1)a+\sqrt{H} \right\} \\
&=-2\left\{ (n-1)^2a^3-2n(2n-3)ab \right\}\sqrt{H}+2\left\{ (n-1)a^2-2nb \right\}H \\
&=-2F\sqrt{H}+2GH, \\[3mm]
&8n^2(n-2)b^2+\left\{ (n-1)a^2-2nb \right\}B^2+4n(n-2)abB \\
&=8n^2(n-2)b^2+\left\{ (n-1)a^2-2nb \right\}\left\{ 2(n-1)^2a^2-4n(n-2)b+2(n-1)a\sqrt{H} \right\} \\
&\hspace{70mm}+4n(n-2)ab\left\{ -(n-1)a-\sqrt{H} \right\} \\
&=2\left\{ (n-1)^2a^3-2n(2n-3)ab \right\}\sqrt{H}+2\left\{ (n-1)a^2-2nb \right\}H, \\
&=2F\sqrt{H}+2GH.
\end{align*}
\normalsize
Hence, by putting
\begin{align*}
A^{n-3}=\left\{-(n-1)a+\sqrt{(n-1)^2a^2-4n(n-2)b} \right\}^{n-3}=I+J\sqrt{H},
\end{align*}
we have
\begin{align*}
&\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2} \\[1mm]
&=\frac{\left\{-2F\sqrt{H}+2GH\right\}\left( I+J\sqrt{H} \right)-\left\{2F\sqrt{H}+2GH\right\}\left( I-J\sqrt{H} \right)}{2^{n-1}n(n-2)^{n-3}\sqrt{H}}t \\
&=\frac{-FI+GHJ}{2^{n-3}n(n-2)^{n-3}}t.
\end{align*}
Then, by using   \cref{lem6.6}, we have
\begin{align*}
I&=\sum_{\ell=0}^{m_0}\binom{n-3}{2\ell}\left\{ -(n-1)a \right\}^{n-2\ell-3}\left\{ (n-1)^2a^2-4n(n-2)b \right\}^{\ell} \\
&=\sum_{\ell=0}^{m_0}\binom{n-3}{2\ell}\left\{ -(n-1)a \right\}^{n-2\ell-3}\left\{\sum_{k=0}^{\ell}\binom{\ell}{k}\left\{ (n-1)^2a^2 \right\}^{\ell-k}\left\{ -4n(n-2)b \right\}^{k}\right\} \\
&=\sum_{k=0}^{m_0}\sum_{\ell=k}^{m_0}\binom{n-3}{2\ell}\binom{\ell}{k}(-1)^{n+k+1}2^{2k}n^k(n-1)^{n-2k-3}(n-2)^ka^{n-2k-3}b^k \\
&=\sum_{k=0}^{m_0}2^{n-2k-4}\left\{\binom{n-k-3}{k}+\binom{n-k-4}{k-1} \right\} \\
&\hspace{43mm}\times (-1)^{n+k+1}2^{2k}n^k(n-1)^{n-2k-3}(n-2)^ka^{n-2k-3}b^k \\
&=\sum_{k=0}^{m_0} (-1)^{n+k+1}2^{n-4}n^k(n-1)^{n-2k-3}(n-2)^k\frac{n-3}{n-k-3}\binom{n-k-3}{k}a^{n-2k-3}b^k
\end{align*}
and
\begin{align*}
J&=\sum_{\ell=0}^{m_0}\binom{n-3}{2\ell+1}\left\{ -(n-1)a \right\}^{n-2\ell-4}\left\{ (n-1)^2a^2-4n(n-2)b \right\}^{\ell} \\
&=\sum_{\ell=0}^{m_0}\binom{n-3}{2\ell+1}\left\{ -(n-1)a \right\}^{n-2\ell-4}\left\{\sum_{k=0}^{\ell}\binom{\ell}{k}\{ (n-1)^2a^2 \}^{\ell-k}\left\{ -4n(n-2)b \right\}^k\right\} \\
&=\sum_{k=0}^{m_0}\sum_{\ell=k}^{m_0}\binom{n-3}{2\ell+1}\binom{\ell}{k}(-1)^{n+k}2^{2k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^k \\
&=\sum_{k=0}^{m_0}2^{n-2k-4}\binom{n-k-4}{k}(-1)^{n+k}2^{2k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^k \\
&=\sum_{k=0}^{m_0}(-1)^{n+k}2^{n-4}n^k(n-1)^{n-2k-4}(n-2)^k\binom{n-k-4}{k}a^{n-2k-4}b^k,
\end{align*}
which implies,

\begin{align*}
&-FI+GHJ \\
&=-\{(n-1)^2a^3-2n(2n-3)ab\}I+\{ (n-1)^3a^4-2n(n-1)(3n-5)a^2b+8n^2(n-2)b^2 \}J \\
&=\sum_{k=0}^{m_0}(-1)^{n+k}2^{n-4}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^k \\
&\hspace{5mm}\times \biggr\{ (n-1)^3\frac{n-3}{n-k-3}\binom{n-k-3}{k}a^4-2n(n-1)(2n-3)\frac{n-3}{n-k-3}\binom{n-k-3}{k}a^2b \\
&\hspace{8mm}+(n-1)^3\binom{n-k-4}{k}a^4-2n(n-1)(3n-5)\binom{n-k-4}{k}a^2b+8n^2(n-2)\binom{n-k-4}{k}b^2 \biggl\} \\
&=\sum_{k=0}^{m_0}(-1)^{n+k}2^{n-3}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^k \\
&\hspace{5mm}\times \biggr\{ (n-1)^3\binom{n-k-3}{k}a^4-\frac{n(n-1)\left\{ 5n^2-(6k+23)n+10k+24 \right\}}{n-k-3}\binom{n-k-3}{k}a^2b \\
&\hspace{88mm}+4n^2(n-2)\binom{n-k-4}{k}b^2 \biggl\}
\end{align*}
\normalsize
Therefore, we finally obtain

\begin{align*}
\alpha(t)&=\left\{\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2}\right\}t^2 =\frac{-FI+GHJ}{2^{n-3}n(n-2)^{n-3}}t^3 \\
&=\sum_{k=0}^{m_0}\frac{(-1)^{n+k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^kS_k}{n(n-2)^{n-3}}t^3,
\end{align*}
\normalsize
where

\begin{align*}
S_k&=(n-1)^3\binom{n-k-3}{k}a^4-\frac{n(n-1)\left\{ 5n^2-(6k+23)n+10k+24 \right\}}{n-k-3}\binom{n-k-3}{k}a^2b \\
&\hspace{81mm}+4n^2(n-2)\binom{n-k-4}{k}b^2,
\end{align*}
\normalsize
which completes the proof of \cref{lem6.5}.
\end{proof}

\begin{proof}[Proof of  \cref{thm6.0}]
Theorem 3 has been proved for $n=3, 4$ and hence let us assume $n \geq 5$. Then, by the definition of $A_{\bm c}(t)_{n-2}$ and equations \eqref{eq6.4}, \eqref{eq6.5}, we have
\begin{align*}
&\Delta\left(f_{\bm c}(t; x)\right)=\det A_{\bm c}(t)=n\cdot\det A_{\bm c}(t)_{n-2} \\
&=n\cdot(-1)^{\lceil (n-3)/2 \rceil}(n-2)^{n-3}t^{n-3}\cdot
\det
\left[
\begin{array}{cc}
a_{n-1,n-1}^{({\bm c})}(t)_{n-2} & a_{n-1,n}^{({\bm c})}(t)_{n-2} \\ \notag
a_{n,n-1}^{({\bm c})}(t)_{n-2} & a_{n,n}^{({\bm c})}(t)_{n-2}
\end{array}
\right].
\end{align*}

Therefore, by  \cref{lem6.5}, we have

\begin{align*}
\Delta\left(f_{\bm c}(t; x)\right)=(-1)^{m_1}t^{n-1}\Biggl\{(n-2)^{n-2}(a^2-4b)t^2+\gamma_{\bm c}t-n^nb^{n-1} \Biggr\},
\end{align*}
where

\begin{align*}
\gamma_{\bm c}=
\begin{cases}
\displaystyle \sum_{k=0}^{m_0}(-1)^{n+k}n^k(n-1)^{n-2k-4}(n-2)^ka^{n-2k-4}b^kS_k & \hspace{-1mm} \text{$(a \neq 0$,  or $a=0$, $n : $ even$)$}, \\
0 & \hspace{-1mm} \text{$(a=0$, $n : $ odd$)$},
\end{cases}
\end{align*}
\normalsize
which completes the proof.
\end{proof}

Now we are ready to prove \cref{thm6.1}.

\begin{proof}[Proof of  \cref{thm6.1}]
Put
$Q(t)=(n-2)^{n-2}(a^2-4b)t^2+\gamma_{\bm c}t-n^nb^{n-1}$
and suppose $b=(n-1)^2a^2/4n(n-2)$.
Then, by equation \eqref{eq6.5} and  \cref{lem6.5}, we have

\begin{align*}
&Q(t) \\
&=n(n-2)^{n-3}\Biggl\{\frac{(n-2)(a^2-4b)}{n}t^2+\left\{\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2}\right\} \\[1mm]
&\hspace{105mm}-\frac{n^{n-1}b^{n-1}}{(n-2)^{n-3}} \Biggr\} \\
&=(n-2)^{n-2}\left\{a^2-\frac{(n-1)^2a^2}{n(n-2)}\right\}t^2 \\
&\hspace{18mm}+n(n-2)^{n-3}\left\{\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2}\right\}-n^nb^{n-1} \\
&=-\frac{(n-2)^{n-3}a^2}{n}t^2 \\
&\hspace{18mm}+n(n-2)^{n-3}\left\{\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2}\right\}-n^nb^{n-1}.
\end{align*}
\normalsize
Here, by equation \eqref{eq6.3}, we have

\begin{align*}
&n(n-2)^{n-3}\left\{\frac{2nb^2\bar{x}_{n-3}}{n-2}+\frac{\left\{(n-1)a^2-2nb\right\}\bar{x}_{n-1}}{n}+2ab\bar{x}_{n-2}\right\} \\
&=n(n-2)^{n-3}\Biggl\{\frac{2n}{n-2}\cdot\frac{(n-1)^4a^4}{16n^2(n-2)^2}\cdot\frac{(-1)^{n-3}(n-3)\{-(n-1)a\}^{n-4}}{2^{n-4}\left\{ -(n-2) \right\}^{n-5}}t \\
&\hspace{30mm}+\left\{\frac{(n-1)a^2}{n}-\frac{(n-1)^2a^2}{2n(n-2)}\right\}\cdot\frac{(-1)^{n-1}(n-1)\{-(n-1)a\}^{n-2}}{2^{n-2}\left\{ -(n-2) \right\}^{n-3}}t \\
&\hspace{59mm}+\frac{(n-1)^2a^3}{2n(n-2)}\cdot\frac{(-1)^{n-2}(n-2)\{-(n-1)a\}^{n-3}}{2^{n-3}\left\{ -(n-2) \right\}^{n-4}}t\Biggr\} \\
&=n(n-2)^{n-3}\left\{\frac{(-1)^n(n-1)^n(n-3)a^n}{2^{n-1}n(n-2)^{n-2}}t+\frac{(-1)^n(n-1)^n(n-3)a^n}{2^{n-1}n(n-2)^{n-2}}t-\frac{(-1)^n(n-1)^{n-1}a^n}{2^{n-2}n(n-2)^{n-4}}t\right\} \\
&=-\frac{(-1)^n(n-1)^{n-1}a^n}{2^{n-2}(n-2)}t,
\end{align*}
\normalsize
which implies
\begin{align*}
Q(t)&=-\frac{(n-2)^{n-3}a^2}{n}t^2-\frac{(-1)^n(n-1)^{n-1}a^n}{2^{n-2}(n-2)}t-n^nb^{n-1}, \\
\Delta\left(Q(t)\right)&=\left\{ -\frac{(-1)^n(n-1)^{n-1}a^n}{2^{n-2}(n-2)} \right\}^2-4\cdot\left\{ -\frac{(n-2)^{n-3}a^2}{n} \right\}\cdot\left( -n^nb^{n-1}\right) \\
&=\frac{(n-1)^{2n-2}a^{2n}}{2^{2n-4}(n-2)^2}-\frac{4(n-2)^{n-3}a^2}{n}\cdot n^{n} \cdot\left\{\frac{(n-1)^2a^2}{4n(n-2)}\right\}^{n-1}=0
\end{align*}
and hence, by completing the square, we have
\begin{align*}
Q(t)=-\frac{(n-2)^{n-3}a^2}{n}\left\{t+\frac{(-1)^nn(n-1)^{n-1}a^{n-2}}{2^{n-1}(n-2)^{n-2}} \right\}^2.
\end{align*}
Thus, if $n$ is even, we have $\alpha_{\bm c}=0$ and hence by \cref{thm5.2}, we have $N_{f_{\bm c}(t; x)}=0$ for any positive real number $t$ since $x^2+ax+b$ has no real root in this case. Then, by considering the graph of the function $y=x^n+t(x^2+ax+b)$ $(t>0)$, we can conclude that $N_{f_{\bm c}(t; x)}=0$ whenever $(n-1)^2a^2/4n(n-2)\leq b$.
\end{proof}

\begin{rem}
It was pointed out to us by J. Gutierrez that the proof of the Cor.~1, it is elementary and all the machinery of \cref{thm6.0} is not needed. However, we decided to present it here with the intention that perhaps such techniques can be generalized to polynomials $f(x)=x^n + t\cdot g(x)$, $\deg g \geq 3$.
\end{rem}

\nocite{*}

\bibliographystyle{amsalpha}

\bibliography{ref}{}

\end{document}


text:
\begin{document}

\title{On  Some Nonstandard  Heat Transfer Problems}
\author{Ashot Djrbashian,  Armen Arakelyan }

\begin{abstract}
In this article we are making an attempt to connect the theory of functions integrable in the unit disk to the problems of steady-state heat transfer in cases when the heat source is inside the disk.

\end{abstract}

\maketitle

\section{Introduction}

Let $f(z)$ be analytic in the unit disk $D=\{z: |z|<1\}$ and for $0<p<\infty, -1<\alpha<\infty$
\begin{equation}
||f||^p_\alpha=\int_D|f(z)|^p(1-|z|)^\alpha dm_2(z)<\infty,
\end{equation}
where $dm_2(z)=dxdy$ is the plane Lebesgue measure. Then we say that $f$ belongs to the class $A^p_\alpha=A^p_\alpha(D)$. The theory of $A^p_\alpha$ spaces was established in 1940's by M.M.Djrbashian (see [3] or Djrbashian, Shamoian [2]) and by now is one of the most advanced branches of the theory of Banach spaces of analytic functions. The spaces $A^p_\alpha$ are natural generalizations of the theory of Hardy spaces $H^p, 0<p\leq\infty$ which itself was establish in 1910's by the efforts of multiple mathematicians. In the groundbreaking paper [3] M.M.Djrbashian proved the following integral representation formula:

\begin{theorem}
Let $f\in A^p_\alpha, \; 1\leq p<\infty, \; -1<\alpha<\infty$. Then for $z=re^{i\theta}, \; w=\rho e^{i\phi}$ we have the following integral representation:
\begin{equation}
f(z)=\frac{1+\alpha}{\pi}\int_0^1\int_{-\pi}^\pi(1-\rho^2)^\alpha\frac{f(\rho e^{i\phi})}{(1-z\rho e^{-i\phi})^{2+\alpha}}\rho d\rho d\phi
\end{equation}
\end{theorem}

As we have mentioned above the mathematical theory of these classes is well known for long time and may be found in the monographs Djrbashian, Shamoian [2], Hadenmalm, Korenblum, Zhu  [6], or Duren, Schuster  [5] (in the special case $\alpha=0$ only). However, unlike the theory of Hardy spaces, as far as we know there are no applications of this theory to concrete physical or engineering problems. In this paper we are doing an attempt to connect the formula (2) in its most basic case to some heat transfer problems. For that purpose we are considering only the simplest case $p=2, \alpha=0$ to make it accessible to widest possible audience of applied mathematicians and engineers with the hope that this approach will promote some additional research. Our idea is that the formula (2) is "too good" to not have useful applications. In some sense we can say, paraphrasing the title of a famous play, that this is  "A formula in search of the problem".

\section{Some integral kernels}
\label{sec:headings}

In this article we will be dealing with harmonic functions  defined in the unit disk $D$. By definition, a function $f(x,y)$ is harmonic if
$$\Delta f(x,y)=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0.$$
In polar coordinates $x=r\cos\theta, y=r\sin\theta$ the above equation has the form
$$\frac{\partial^2 f}{\partial r^2}+\frac{1}{r}\frac{\partial f}{\partial r}+\frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2}=0.$$

Poisson kernel for the unit disk $D$ is the function (in polar coordinates)
\begin{equation}
 P_r(\theta)=P(r,\theta)=\frac{1-r^2}{1-2r\cos\theta+r^2}.
 \end{equation}
It is a long established fact that this kernel solves the so-called classical Dirichlet problem for Laplace's equation in the disk: Find a harmonic function $u$ inside the disk  that on the boundary $T=\partial D$ has the prescribed continuous values $f$. The solution is just the convolution of that function $f$ and the Poisson kernel:
\begin{equation}
u(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^{ \pi}f(\varphi)P_r(\theta-\varphi)d\varphi.
\end{equation}
This solution in effect says that in order to know the values of any harmonic function inside the disk it is sufficient to know its values on the circle, the boundary of the disk.

In fact this formula extends to much wider class of boundary functions $f(\theta)$ than continuous functions. For example, if $[a,b]\subset [-\pi,\pi]$ and $f$ is the characteristic function of that interval then formula (2) still produces a harmonic function $u$ that has boundary values equal to $f$ (in some more general sense). For our purposes we will assume  that $f\in L^2(T)$, the Lebesgue space of square integrable  functions on the unit circle. The resulting function $u$ will belong to harmonic Hardy class $h^2$ satisfying the condition

$$\sup_{0\leq r<1}\int_{-\pi}^\pi |f(re^{i\theta})|^2 f\theta<\infty.$$

The Poisson kernel (3) is one of many examples of  "reproducing kernels". Among the most important properties of the Poisson kernel is the fact that  $P_r(\theta)\geq0$ and for any $r, 0\leq r<1, \frac{1}{2\pi}\int_{-\pi}^{\pi}P_r(\theta)d\theta=1.$ Below you can see the graph of the kernel $P$ for the values  $r=0.5, 0.75,$ and  $0.85$.

\centerline{\includegraphics[width=4in]{Poisson.jpg}}

$$Figure \; 1$$

The term reproducing kernel is used to indicate the property of reproducing a given harmonic function from its boundary values. We will return to this discussion a little later when we introduce the next integral kernel for harmonic functions.

All the facts about Poisson kernels and Poisson integrals are so widely known that could be found in virtually any Complex Analysis or PDE textbook. We refer interested reader to the book  by Duren [4], for example.

Next let us consider a different type of integral kernel and integral representation for harmonic functions in the disc. Here the functions are taken from much wider class satisfying condition

\begin{equation}
    \int_{D}|u(x,y|^2dxdy<\infty.
\end{equation}

Here integration is over the unit disk with respect of regular plane Lebesgue measure and we will denote the class of such functions by $hA^2=hA^2(D)$. It is widely known that unlike harmonic functions from the Hardy class $h^2$ functions from $hA^2$ generally speaking do not have boundary values. That means (again, generally speaking) that no representation with Poisson kernel is possible for functions of the class $hA^2$. For these and many other facts about these classes see, e.g. Djrbashian , Shamoian [2] or Axler, Bourdon, Ramey  [1]. The following theorem was never formulated as a separate result anywhere but can be easily deduced from much more general theorems 1.2 or 7.1 from [2].
\begin{theorem}
If $u\in hA^2$ then it has the integral representation
\begin{equation}
u(r,\theta)=-u(0)+\frac{2}{\pi}\int_0^1\int_0^{2\pi}u(\rho,\phi)\frac{1-2r\rho\cos(\theta-\phi)+r^2\rho^2\cos2(\theta-\phi)}{(1-2r\rho\cos(\theta-\phi)+r^2\rho^2)^2}d\phi \rho d\rho
\end{equation}
\end{theorem}

Even though the kernel function

\begin{equation}
  Q(r,\theta)=  \frac{1-2r\cos\theta+r^2\cos2\theta}{(1-2r\cos\theta+r^2)^2}
\end{equation}
does not formally reproduce the function $u(r,\theta)$ (we still need to subtract the value of $u$ at the origin) we will call it  reproducing kernel for the space $hA^2$ and in the future disregard the first term in formula (6) unless it is absolutely necessary.

By taking $u(r,\theta)=1, 0\leq r<1, 0\leq\theta<2\pi$ we easily see that

$$\frac{1}{\pi}\int_0^1\int_0^{2\pi}Q(r\rho,\theta-\phi)\rho d\theta d\rho =1$$
for all values of $r, \theta$.

It is also important to notice that in order to "recover" any given harmonic function from that class we need the values of that function on every point in the disk. That is the crucial difference between Poisson representation (4) and representation with the kernel $Q$.

In what follows we will establish some important properties of the kernel function $Q(r,\theta)$. First of all notice that it is not non-negative and that will be demonstrated below with the graph of the function for different values of $r$.  However many important properties are still valid here, starting of course with its reproducing kernel property, expressed by the formula (6).

We start with the graphs of the kernel $Q$ for the values $r=0.5, 0.75$

\centerline{\includegraphics[width=4in]{Q-Kernel.jpg}}

$$Figure \; 2$$

As $r$ gets closer to 1 the graph becomes narrower and the peak higher and, not surprisingly, tends to Dirac $\delta$-function. In that sense the $Q$-kernel behaves almost identically with the Poisson kernel except of the fact that it is not non-negative. Before going to differences here we would like to demonstrate more similarities between the two integral representation formulas (4) and (6). Let us consider one of the simplest possible harmonic functions $u=x=r\cos\theta$. The boundary value of this function on the unit circle is obviously $f(\theta)=\cos\theta$ (here and later we write $f(\theta)$ instead of $f(e^{i\theta})$, i.e. we identify the unit circle with the interval $[0,2\pi]$). Because the function $u$ is harmonic we would obviously have
$$r\cos\theta=\frac{1}{2\pi}\int_{-\pi}^{ \pi}\cos\phi \frac{1-r^2}{1-2r\cos(\theta-\phi)+r^2} d\phi$$
and (because $u(0)=0),$
$$r\cos\theta=\frac{2}{\pi}\int_0^1\int_0^{2\pi}\rho\cos\phi\frac{1-2r\rho\cos(\theta-\phi)+r^2\rho^2\cos2(\theta-\phi)}{(1-2r\rho\cos(\theta-\phi)+r^2\rho^2)^2}d\phi \rho d\rho.$$

In other words, formulas (4) and (6) produce (reconstruct) the exact same function. This would be the case any time we use these formulas for functions harmonic in the unit disk. We will demonstrate this fact also graphically below in the case of another simple harmonic function $u(x,y)=x^2-y^2=r^2\cos2\theta$ with the boundary value $f(\theta)=\cos2\theta$.

\centerline{\includegraphics[width=6in]{Fig1.jpg}}

$$Figure \; 3$$

As it is clearly seen the graphs match perfectly not just in shape and form but also numerically.\footnote{Numeric outputs for these and all the following graphs are available upon request by writing to the first author}

Next we will start concentrating on essential differences between these two integral representations and in order to do that we will need some more general results about the kernel $Q$ and the integral formula (6). For that purpose it is convenient to consider an integral operator

\begin{equation}
Tf(r,\theta)=\int_D f(\rho,\phi)Q(r\rho,\theta-\phi)\rho d\rho d\phi
\end{equation}
where the function $f$ belongs to the Lebesgue space $L^2=L^2(D, dxdy).$ Then the following result is a very special case of the Theorem 7.3 from [2]:

\begin{theorem}
The operator $T$ is a bounded projection from $L^2$ to $hA^2$ and the norm of this operator is $\leq1$:  $||Tf||_{hA^2}\leq||f||_{L^2}$
\end{theorem}

This means, in particular, that for any square integrable function $f$ (harmonic or not) the result of its convolution with the kernel $Q$ is a harmonic function which belongs to the class $hA^2$. Calculating integral (8) for non-harmonic functions explicitly is very difficult except when $f$ is a polynomial. For a software allowing to do it symbolically using Mathematca see [1]. For any other types of functions, especially non-continuous functions, it is basically impossible. In this article we, for the first time,  have collected a number of cases where we have calculated this integral numerically and graphed them.

Our first example will be the $Q$-integral of the characteristic function of the disk with radius $r=1/4$ centered at the origin.
\begin{equation}
    \int_0^{1/4}\int_0^{2\pi}Q(r\rho,\theta-\phi)\rho d\phi d\rho.
\end{equation}
The output function is depicted below:

\centerline{\includegraphics[width=4in]{Fig3.jpg}}

$$Figure \;4$$

The height of the disk is about $\pi/16\approx 0.193.$ This result seems a little surprising but makes sense because the resulting  function should be harmonic and also makes sense from a physical point of view. We will return to this issue later in the next section.

In the next two examples we first calculate the $Q$-integral of the characteristic function of the polar rectangle $[1/4,1/2]\times[0,\pi/4]$
$$\int_{1/4}^{1/2}\int_0^{\pi/4}Q(r\rho,\theta-\phi)\rho d\phi d\rho.$$
and the next graph is the $Q$-integral of the above characteristic function plus the characteristic function of the polar rectangle $[0.6,0.8]\times[5\pi/6, \pi]$:
$$\int_{1/4}^{1/2}\int_0^{\pi/4}Q(r\rho,\theta-\phi)\rho d\phi d\rho+ \int_{0.6}^{0.8}\int_{5\pi/6}^{\pi}Q(r\rho,\theta-\phi)\rho d\phi d\rho.$$
The graphs are presented below side by side for comparison.

\centerline{\includegraphics[width=6in]{Fig4.jpg}}

$$Figure \; 5$$

By Theorem 2 both functions are clearly harmonic. It is also obvious from comparison that the second function has more "mass" because the input is bigger. If the highest point of the first function is just about 0.17 then for the second one it is about 0.5.

Our next example is a function that goes to infinity near the boundary of the unit disk. The function
$$f(r,\theta)=\frac{\cos\theta}{(1-r)^{1/4}}$$
is not harmonic but clearly belongs to $L^2(D)$. We take the $Q$-transform of that function multiplied by the characteristic function of the "rectangle" $[3/4,1]\times[-\pi/6,\pi/6]$:
$$\int_{3/4}^1\int_{-\pi/6}^{\pi/6}\frac{\cos\phi}{(1-\rho)^{1/4}}Q(r\rho,\theta-\phi)\rho d\phi d\rho.$$

\centerline{\includegraphics[width=4in]{Fig5.jpg}}

$$Figure \; 6$$

The resulting function grows near the point $\theta=0$ and the picture is necessarily cut off because of that.

And in our final example we would like to show the behaviour of the $Q$-transform of another $L^2$ function going to infinity near the boundary which is the combination of the previous integral and a similar function that tends to $-\infty$ on the opposite side of the disk:

$$\int_{3/4}^1\int_{-\pi/6}^{\pi/6}\frac{\cos\phi}{(1-\rho)^{1/4}}Q(r\rho,\theta-\phi)\rho d\phi d\rho+\int_{7/8}^1\int_{5\pi/6}^{\pi}\frac{\cos\phi}{(1-\rho)^{3/8}}Q(r\rho,\theta-\phi)\rho d\phi d\rho.$$

You can see from the figure bellow that the behaviour of this graph almost completely matches the behaviour of the previous one near $\theta=0$. On the other hand on around the interval $[5\pi/6,\pi]$ it has the complete opposite behaviour. As in the previous graph we had to cut off the graph to allow the program work.

\centerline{\includegraphics[width=4in]{Fig6.jpg}}

$$Figure \; 7$$

\section{The steady-state heat equation}

Let us return to the classical Dirichlet problem for the Laplace's equation in the disk discussed at the beginning of the previous section:
 \begin{equation}
 \Delta u=0, \quad u|_{T}=f,
 \end{equation}
where $f\in C(T)$. As we have mentioned earlier the solution to this boundary problem is given by the Poisson integral
\begin{equation}
    u(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\phi)\frac{1-r^2}{1-2r\cos(\theta-\phi)+r^2}d\phi.
\end{equation}
However, the formula (11) is valid for much larger class of functions $f(\theta)$ but we restricted our interest to the case $f\in L^2(T)$ to avoid discussion of fine properties of harmonic functions at the boundary. Among other things the boundary problem (10) solves the following "steady-state" (or time independent) heat transfer problem: Assume that we establish and maintain certain temperature on the unit circle which is described by the function $f(\theta)$. How the temperature will be distributed at each point of the disk. Formula (11) gives exact answer to that question. It is also important to remind that the resulting function $u(r,\theta)$ is harmonic and, in particular, satisfies the minimum and maximum principles for harmonic functions: Function $u$ can reach its maximum or minimum values only on the boundary or that function is constant.

Even though the results described above (and even their very far reaching generalizations) are known for a very long time we find it useful to provide some illustrations of behaviour of solutions of the boundary problem (10).

In our first example the function $f$ is just the characteristic function of the integral $[-\pi/6,\pi/6]$ (otherwise called the harmonic measure of that interval):

$$  u(r,\theta)=\frac{1}{2\pi}\int_{-\pi/6}^{\pi/6}\frac{1-r^2}{1-2r\cos(\theta-\phi)+r^2}d\phi$$

\centerline{\includegraphics[width=4in]{Fig21.jpg}}

$$Figure \; 8$$

As we clearly see the resulting function reaches its maximum value 1 on some interval around $\theta=0$ and is zero on the compliment of that interval. Disregarding some complications at the end points $\pm \pi/6$ inside the disk the resulting function is clearly harmonic.

Now let us consider a different physical problem. Suppose we are heating not just boundary (or part) of the disk but a whole layer close to the boundary. For example, that might be a polar rectangle of the width 0.1 $R=[0.9,1]\times[-\pi/6,\pi/6]$. It is obvious that the Poisson formula (11) will not work in this case. It is also clear from physical considerations that the difference should not be crucial. So, let us see if the $Q$-transform can do the job here. In other words, let us calculate and graph the function
\begin{equation}
\frac{2}{\pi}\int_{0.9}^1\int_{-\pi/6}^{\pi/6}\frac{1-2r\rho\cos(\theta-\phi)+r^2\rho^2\cos2(\theta-\phi)}{(1-2r\rho\cos(\theta-\phi)+r^2\rho^2)^2}d\phi \rho d\rho.
\end {equation}

The resulting function is depicted in the graph below from two different angles:

\centerline{\includegraphics[width=5in]{Fig22.jpg}}

$$Figure \; 9$$

Let us now compare the form and numeric values of functions in Figures 8 and 9.  First of all we notice that the shape of the graphs are almost identical away from "critical" rectangle $R$. However, the numeric values of the $Q$-integral are just about half of the Poisson integral. When we approach to the rectangle $R$ differences are becoming even more significant and near the very edge a big drop in values happens: at some small parts in $R$ the values even become negative. That is clearly visible in the  right-hand side picture in Figure 9.

So, the first conclusion seems to be that this approach to the solution of the "non-standard" problem is unlikely to be given by the integral (12). However, let us continue with more examples and compare the resulting graphs.In our second example the function $f$ in (11) is $|\theta|$ and we present it in two different views:

\centerline{\includegraphics[width=6in]{Fig23.jpg}}

$$Figure \; 10$$

Next we will try to pose the corresponding question similar to the previous case. Suppose a whole layer of the disk $D$ is heated with the width 0.1, say. What will be the heat distribution throughout the whole disk. Let us try again to use the $Q$-transform:

$$\frac{2}{\pi}\int_{0.9}^1\int_{-\pi}^{\pi}\frac{1-2r\rho\cos(\theta-\phi)+r^2\rho^2\cos2(\theta-\phi)}{(1-2r\rho\cos(\theta-\phi)+r^2\rho^2)^2}|\phi|\rho^2d\phi d\rho.$$

The resulting graphs are below:

\centerline{\includegraphics[width=6in]{Fig24.jpg}}

$$Figure \; 11$$

Comparison of Figures 10 and 11 gives us a much better match. The graphs are clearly very similar (although certainly not identical). The numeric values in the second case are again roughly the half of the first one.

Generally speaking, when comparing Poisson integrals of certain functions on the unit circle and $Q$-transforms of corresponding functions on a narrow strip about the circle we see that sometimes there are substantial differences and in many other cases general similarities in shapes of corresponding graphs. In both situations, of course, numerically they are different but seems that follow the same pattern: Poisson integrals (at least for positive functions) are greater than the $Q$-transforms. To confirm this assertion we will present some more examples.  In the next graph we put next to each other Poisson integral of the function $f(\theta)=\theta^2$ on interval $[-\pi/6,\pi/6]$ and the $Q$-integral of the corresponding function $f(r,\theta)=r\theta^2$ on rectangle $[0.9,1]\times[-\pi/6,\pi/6]$:
$$\frac{1}{2\pi}\int_{-\pi/6}^{\pi/6}P_r(\theta-\phi)\phi^2 d\phi, \quad \frac{2}{\pi}\int_{0.9}^1\int_{-\pi/6}^{\pi/6} Q(r\rho, \theta-\phi)\phi^2\rho^2 d\phi d\rho$$
The resulting graphs are below:

\centerline{\includegraphics[width=6in]{Fig25.jpg}}

$$Figure \; 12$$

Here we clearly see the similarities of the behaviour of both harmonic functions and the differences in numeric values. We can see a very similar pattern also in the next example where we have taken the Poisson integral of the function $\sin\theta$ over interval $[0,\pi$ and the $Q$-transform of the corresponding function $r\sin\theta$ over the rectangle $[0.9,1]\times[0,\pi]$:

$$\frac{1}{2\pi}\int_0^{\pi}P_r(\theta-\phi)\sin\phi d\phi, \quad \frac{2}{\pi}\int_{0.9}^1\int_0^{\pi} Q(r\rho, \theta-\phi)\sin\phi\rho^2 d\phi d\rho$$

\centerline{\includegraphics[width=6in]{Fig26.jpg}}

$$Figure \; 13$$

And again we see pretty similar pattern except the fact that the second function takes some negative values too near the critical points $\theta=0, \pi.$

Our last example in this part is a function with integrable singularity, namely the logarithmic function:

$$\frac{1}{2\pi}\int_0^{\pi}P_r(\theta-\phi)|\ln|\phi|| d\phi, \quad \frac{2}{\pi}\int_{0.9}^1\int_0^{\pi} Q(r\rho, \theta-\phi)|\ln|\phi||\rho^2 d\phi d\rho$$
and the graphs  below again show a similar pattern as in the previous examples:

\centerline{\includegraphics[width=6in]{Fig27.jpg}}

$$Figure \; 14$$

At this point we would like to return to some examples from Section 2. In these examples comparison with Poisson integrals is no longer possible because the functions we have used there were supported inside the disk either completely or had significant "mass" inside the unit disk.

Let us look back at the integral (9) and the corresponding graph in Figure 4 first. Then we can see that the result conforms (at least in general terms) with physical intuition. Really, in that integral if we view the characteristic function of the disk $r\leq 1/4$ as a source of heat of "intensity" one, then the common sense tells us that eventually  the heat will be spread uniformly throughout the whole disk $r\leq1$ but the temperature will be lower than initial temperature 1. Pay attention that in this case it would be impossible to apply Poisson integral to come to the same conclusion.

We will come to similar conclusions if we look at integrals that resulted in graphs depicted in Figures 5, 6, and 7. It could be claimed with reasonable certainty  that these graphs represent the heat distribution in the disk with the corresponding  "heat sources".

We would like to add one more example along these lines. The function which we view as a heat source is $f(r,\theta)=10e^{-10(r-0.5)^2}\cos\theta$ restricted to the "rectangle" $[0.3,0.7]\times[-\pi/6,\pi/6]$. This function is supported inside the disk and, again, cannot be handled with Poisson integral. The resulting graph represents a harmonic function that has growth tendency in the direction of the boundary where the presumed heat source is concentrated. We present below that graph along with the function $f(r,\theta)$ to the right.

\centerline{\includegraphics[width=6in]{Fig28.jpg}}

$$Figure \; 15$$

Based on all the previous examples we would like to formulate a plausible conjecture:\\

\noindent\textbf {Conjecture} {\it Suppose a metallic disk is getting  heated by a source described by the function $f(r,\theta)$. Then the steady-state heat distribution through the disk, assuming there are no outside factors, is given by the integral (8) modulo a multiplicative constant depending on physical properties of the disk.} \\

 Mathematical proof seems to be elusive at this juncture but it could be confirmed or rejected also experimentally.\\

\noindent CONCLUDING REMARK. The coding, numeric calculations, and graphing part of this article was done completely by the second author who is a student at Glendale Community College.\\

\noindent REFERENCES\\[10pt]

\noindent 1. S.Axler , P. Bourdon, W.Ramey,  "Harmonic Function Theory", Springer Verlag, New York, Berlin, 1992\\
2. A.Djrbashian, F.Shamoian, "Topics in the Theory $A_\alpha^p$ Spaces", Teubner Verlag, Leipzig, 1988\\
3. M.M.Djrbashian, "On the problem of representing analytic functions", Soobshch. Inst.Math. Mech.Acad.Nauk Arm.SSR, N2(1948), 3-40.\\
4. P.L.Duren, "Theory of $H^p$ Spaces", Academic Press 1970\\
5. P.L.Duren, A.Schuster, "Bergman Spaces", AMS, Providence RI, 2004\\
6. H.Hadenmalm, B.Korenblum, K.Zhu, "Theory of Bergman Spaces", Springer Verlag, New York, Berlin, 2000\\

Ashot Djrbashian

Glendale Community College, Glendale CA

E-mail: ashotd@glendale.edu

Armen Arakelyan

Glendale Community College, Glendale, CA

E-mail: aarakel632@student.glendale.edu

\end{document}

config: arxiv


TITLE: Random parking problem on a probability distribution
QUESTION [2 upvotes]: Rényi's Parking Constants comes up when one puts down unit length cars on a interval, such that the probability of covering any two interval is the same.
Are there any published results when the distribution is non-uniform?

REPLY [6 votes]: Here is one: Jean-François Marckert, Parking with density, Random Structures and Algorithms 18 (4), 364-380 (2001). A preprint version is available on this page.


TITLE: How to find $\lim\limits_{t \to \infty} \int\limits_0^\infty u(x,t) dx$?
QUESTION [2 upvotes]: I have $u(x,t)$ - solution of Cauchy problem $$u_t=u_{xx},~~~u(x,0)=e^{-x^2},$$ where $t>0, x\in\mathbb{R}.$
Is there a way to find such a limit? I have seen several theorems which help to find value of $\lim\limits_{t \to \infty} u(x,t),$ but this integral realy spoils everything.
Thank you.

REPLY [2 votes]: The $L_p$ solution to the problem is though convolution with Gauss-Wierstrass kernel,
$$
u(x,t)= W_t*f(x)=\frac{1}{\sqrt{4\pi t}}\int f(y)e^{\frac{-|x-y|^2}{4t}}\,dy
$$
where $f(x)=e^{-x^2}$. The kernel $W_t$ is a good approximation to the identity and so, $W_t*f\rightarrow f$ in $L_p$ ($1\leq p<\infty$ ) and point wise to $f$ (it should be at any Lebesgue point of $f$ fut $f(x)=e^{-x^2}$ is continuous so every point in $\mathbb{R}$ is a Lebesgue point of $f$. In fact, the problem is simple here since $f(x)$ is uniformly continuous.
Here are some generalities of approximation to identity. You have to show that $W_t(x)=t^{-1}W_1(x/t)$, where $W_1(y)=\frac{1}{\sqrt{4\pi}}e^{-y^2}$ satisfies the conditions (1)--(3) described below:

Consider a collection $\{K_\varepsilon:\varepsilon>0\}\subset\mathcal{L}_1(\mathbb{R}^n,\lambda_n)$ that satisfy the following properties:

$\int_{\mathbb{R}^n} K_\varepsilon(x)\,dx=a$ for all
$\varepsilon>0$.
$\sup_{\varepsilon>0}\|K_\varepsilon\|_1<\infty$.
$\int_{|x|>\delta}|K_\varepsilon(x)|\,dx\rightarrow0$ as
$\varepsilon\rightarrow0$.

Theorem.
  Suppose $\{K_\varepsilon:\varepsilon>0\}\subset\mathcal{L}_1(\mathbb{R}^n,\lambda_n)$ satisfy (1)--(3) above. Then, for any  $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, $1\leq p<\infty$,
\begin{aligned}
\lim_{\varepsilon\rightarrow0}\|f* K_{\varepsilon} - a\,f\|_p = 0.
\end{aligned}
If $f\in\mathcal{L}_\infty(\mathbb{R}^n,\lambda_n)$ is continuous at a point $x$, then
$\lim_{\varepsilon\rightarrow0}f*K_\varepsilon(x)=f(x)$. If  $f$ is bounded and uniformly continuous, then $f*K_\varepsilon$ converges to $f$ uniformly as $\varepsilon\rightarrow0$.
Proof:
Let $M=\sup_{\varepsilon>0}\|K_\varepsilon\|_1$. If $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, $1\leq p<\infty$, and application of the generalized Minkowski's inequality gives
\begin{aligned}
\|f* K_\varepsilon -af\|_p &\leq  \Big(\int\Big(\int_{\mathbb{R}^n}|f(x-y)-f(x)|
|K_\varepsilon(y)|\,dy\Big)^p\,dx\Big)^{1/p}\\
& \leq \int_{\mathbb{R}^d}\Big(\int|f(x-y)-f(x)|^p\,dx\Big)^{1/p}|K_\varepsilon(y)|\,dy\\
&=\int\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy.
\end{aligned}
where $\tau_y$ is the translation operator. Recall that $\lim_{y\rightarrow0}\|\tau_yf-f\|_p=0$ for all $f\in\mathcal{L}_p$. Assumption (2) implies that for any $\eta>0$,  there exists $\delta>0$ such that $M\|\tau_yf-f\|_p<\eta/2$ whenever $|y|\leq\delta$. By assumption (3), for some $\varepsilon'>0$ we have
$2\|f\|_p\int_{|y|>\delta}|K_\varepsilon(x)|\,dx<\eta/2$ whenever $\varepsilon<\varepsilon'$. Combining these facts,  we obtain
\begin{aligned}
\|f*K_\varepsilon-af\|_p\leq&
\int_{|y|\leq\delta}\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy\\
&\quad + \int_{|y|>\delta}\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy
\leq \frac{\eta}{2} + \frac{\eta}{2}
\end{aligned}
whenever $0<\varepsilon<\varepsilon'$.
The second statement follows similarly. Let $\eta>0$ be fixed. If $f$ is continuous at $x$, then for some $\delta>0$, $|x-u|\leq\delta$ implies that
$|f(x)-f(u)|<\frac{\eta}{2 M}.$
 For such $\delta>0$, there is $\varepsilon'>0$ such that
$2\|f\|_{\infty}\int_{|x|>\delta}|K_\varepsilon(x)|\,dx<\frac{\eta}{2}$ whenever $0<\varepsilon<\varepsilon'$. Putting these statements together gives
\begin{aligned}
|f*K_\varepsilon(x)&-af(x)|\leq \int |f(x-y)-f(x)||K_\varepsilon|(y)\,dy\\
&\leq
\int_{|y|\leq\delta}+\int_{|y|>\delta}|f(x-y)-f(x)||K_\varepsilon|(y)\,dy
\leq \frac{\eta}{2}+\frac{\eta}{2}
\end{aligned}
If $f$ is bounded and uniformly continuous, then $\delta>0$ can be chosen so that
$$\sup_{|v-u|<\delta}|f(u)-f(u)|<\frac{\eta}{2M}.$$
Uniform convergence follows immediately.


TITLE: Can we add Lagrangian-like multipliers for a joint constraint $g(x, y)=0$ in a min-max problem?
QUESTION [2 upvotes]: Assume we have a problem
$$ \min_x \max_y f(x, y) \ \ \text{ s.t. } \ g(x, y) = 0$$
I wonder if it is possible to incorporate this constraint into an objective in Lagrangian-like fashion:
$$ \min_{x} \max_{y, \lambda} \min_{\eta} f(x, y) + \eta \cdot \lambda \cdot g(x,y) $$
but my attempts at getting such an expression failed so far. For example, if we consider $h(x,y,\lambda) = \min_{\eta} f(x, y) + \eta\lambda g(x,y)$, it already takes the $-\infty$ for all values of $(x, y, \lambda)$ such that $g(x, y) \neq 0$ including $\lambda = \pm \infty $, so there is no reason for $\min_x$ not to choose an $x$ s.t. $g(x,y)\neq0$.
It is no suprrise since I was trying to get some characteristic term $ f(x, y) + \chi(g)(x, y)$ that would be "not favorable" to nor minimization or maximization, so it should not be one of $\pm \infty$.
If constraints are not joint, i.e. $g(x)=0$ and $k(y)=0$ then adding two multiplier terms seems to work $p(x,y,\lambda,\eta) = \max_\eta \min_\lambda f(x, y) + \eta \cdot g(x) + \lambda \cdot k(y)$ is equal to $+\infty$ if $k(y) \neq 0$ and $-\infty$ if $g(x)\neq 0$ and $\infty$ if both do not hold (which is weird, but reasonable). But this is not the case I am interested in.

REPLY [0 votes]: When solving this kind of problem
$$
 \min_x \max_y f(x, y) \ \ \text{ s.t. } \ g(x, y) = 0
$$
we are searching for the $f_g(x,y)$ saddle points and those points are handled normally in the Lagrange Multipliers procedures so
$$
L(x,y,\mu) = f(x,y)+\mu g(x,y)
$$
and the stationary points are determined by
$$
\nabla L = 0
$$
regarding only saddle points.