A class of completely integrable quantum systems associated with classical root systems
Toshio Oshima
We classify the completely integrable systems associated with classical root systems whose potential functions are meromorphic at an infinite point.
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\begin{document}
\title[completely integrable quantum systems]%
{A class of completely integrable quantum systems associated with 
classical root systems}%
\author{Toshio Oshima}
\address{Graduate School of Mathematical Sciences,
University of Tokyo, 7-3-1, Komaba, Meguro-ku, Tokyo 153-8914, Japan}
\email{oshima@ms.u-tokyo.ac.jp}
\dedicatory{Dedicated to Professor Gerrit van Dijk on the occasion of 
his sixty fifth birthday}
\begin{abstract}
We classify the completely integrable systems associated with
classical root systems whose potential functions are meromorphic at an 
infinite point.
\end{abstract}
\maketitle
\section{Introduction}
A Shr\"odinger operator 
\begin{equation}
  P=\sum_{j=1}^n\frac{\p^2}{\p x_j^2} + R(x)\label{eq:Shr}
\end{equation}
with the potential function $R(x)$ of $n$ variables $x=(x_1,\dots,x_n)$ 
is called {\it completely integrable\/} if
there exist $n$ differential operators $P_1,\dots,P_n$ such that
\begin{equation}\label{eq:CI}
 \begin{cases}
   [P_i,P_j]=0\qquad(1\le i < j\le n),\\
   P\in\mathbb C[P_1,\dots,P_n],\\
   P_1,\dots,P_n\text{ are algebraically independent}.
 \end{cases}
\end{equation}
In this note $P$ is called to be completely integrable {\it of type $B_n$\/} 
or {\it of classical type\/}
if $P_k$ and $R(x)$ in the above are of the forms
\begin{align}
 P_k &= \sum_{j=1}^n\frac{\p ^{2k}}{\p x_j^{2k}}+ Q_k\qquad
 \text{with }\ord Q_k<\ord P_k,\label{eq:Bn}\\
 R(x) &= \sum_{1\le i<j\le n}\Bigl(u^-_{ij}(x_i-x_j)+u^+_{ij}(x_i+x_j)\Bigr)
        +\sum_{k=1}^n v_k(x_k).\label{eq:1var}
\end{align}
Here $u^\pm_{ij}$ and $v_k$ are functions of one variable.

The systems of differential operators satisfied by 
the radial parts of zonal spherical functions or Whittaker functions on 
Riemannian symmetric spaces of the non-compact and classical type, 
Heckman-Opdam's hypergeometric equations (cf.~\cite{HO}), 
 Calogero-Moser and Sutherland systems for
one dimensional quantum $n$-body problems (cf.~\cite{OP1}, \cite{OP2})
and Toda finite chains associated with (extended) classical Dynkin diagrams
are their examples.

We remark that \cite{Wa} proves that if the potential function 
$R(x)$ is locally defined and analytic, then the condition \eqref{eq:CI} 
with \eqref{eq:Bn} assures \eqref{eq:1var}
and moreover $R(x)$ is extended to a global meromorphic function on 
$\mathbb C^n$ except for a trivial case corresponding to Type $A_1$ in 
Theorem~\ref{prop:bn}
(cf.\ \cite{Oc} for type $B_2$ and \cite{OS} in the invariant case).

\cite{OOS}, \cite{OS}, \cite{O2} and \cite{OO} determine this integrable 
system under the condition that $P$ is $B_n$-invariant, 
namely, $u^+_{ij}$, $u^-_{ij}$ and $v_k$ are even functions and do not
depend on $i$, $j$ and $k$ and $u^+_{ij}=u^-_{ij}$. 
On the other hand
\cite{Oc}, \cite{Ta} and \cite{Wa} determine it if $R(x)$ has certain
singularities.

We assume in this note
that $R(x)$ is meromorphic at $t=0$ under the coordinate system
\begin{equation}\label{eq:coord}
 t_j = e^{-(x_j-x_{j+1})}\quad(j=1,\ldots,n-1),\quad t_n = e^{-x_n}
\end{equation}
and classify the Shr\"odinger operator \eqref{eq:Shr} 
which allows a differential operator $P_2$ of the form \eqref{eq:Bn} satisfying
$PP_2= P_2P$.
We note that the above examples with non-rational potential functions
satisfy this assumption.
In the first example this follows from the fact that the 
invariant differential operators on a Riemannian symmetric space have 
analytic extensions on a smooth compactification of the space
(cf.~\cite{O1}).

Theorem~\ref{prop:bn} and Remark~\ref{rm:bn} in \S\ref{sec:Bn} determine 
$R(x)$, which is the main result of this note and proved by using
\S\ref{sec:An} and \S\ref{sec:B2}.
The result implies that the system is a suitable
limit of the invariant quantum integrable system classified by \cite{OOS}
(cf.~\cite{I}, \cite{IM}, \cite{Ru}, \cite{vD} and \cite{vD2}).
Hence the integrals $P_1,\dots,P_n$ will be calculated as suitable
limits from the integrals %of the invariant case 
given in \cite{O2}, which 
will be shown in another paper.

If $R(x)$ is analytic at $t=0$, we say that $R(x)$ has 
{\it regular singularity\/} at the infinite point $t=0$, which are also
classified in Corollary~\ref{cor:regB}.

In \S\ref{sec:B2} the potential function $R(x)$ is determined when $n=2$.

In \S\ref{sec:An} we study the potential function $R(x)$ when 
$u^+_{ij}=v_k=0$, which we call to be of type $A_{n-1}$.

\section{Type $A_{n-1}$ ($n\ge3$)}\label{sec:An}
In this section we study the Shr\"odinger operator
\begin{equation}
 P=\sum_{j=1}^n\frac{\p^2}{\p x_j^2} + \sum_{1\le i<j\le n} 
 \tilde u_{ij}(x_i-x_j)
\end{equation}
which allows a differential operator
\begin{equation}
 Q=\sum_{1\le i<i<k\le n}\frac{\p^3}{\p x_i\p x_j\p x_k}+S
 \qquad\text{with }\ord S<3
\end{equation}
satisfying $[P,Q]=[\sum_{j=1}^n\frac{\p}{\p x_j},Q]=0$.
Then the proof of \cite[Proposition 4.2]{OOS} implies that the
existence of $Q$ is equivalent to 
\begin{equation}\label{eq:an}
  \sum_{1\le i<j<k\le n}U_{ijk} = 0
\end{equation}
with
\begin{align}
\begin{aligned}
  U_{ijk} &= 
     u_{jk}(t_j\cdots t_{k-1})\Bigl(t_i\cdots t_{j-1}u'_{ij}(t_i\cdots t_{j-1})
     + t_i\cdots t_{k-1}u'_{ik}(t_i\cdots t_{k-1})\Bigr)\\
     &+ u_{ik}(t_i\cdots t_{k-1})\Bigl(-t_i\cdots t_{j-1}u'_{ij}(t_i\cdots t_{j-1})
     + t_j\cdots t_{k-1}u'_{jk}(t_j\cdots t_{k-1})\Bigr)\\
     &-u_{ij}(t_i\cdots t_{j-1})\Bigl(t_i\cdots t_{k-1}u'_{ik}(t_i\cdots t_{k-1})
     + t_j\cdots t_{k-1}u'_{jk}(t_j\cdots t_{k-1})\Bigr)
\end{aligned}
\end{align}
by putting $u_{ij}(e^{-y}) = \tilde u_{ij}(y)$.
We assume that $R(x)$ is holomorphic for $0<|t|\ll1$ under the coordinate system \eqref{eq:coord} which corresponds to the expression
\begin{equation}
   u_{ij}(s) = \sum_{\nu\in\mathbb Z}c^{ij}_\nu s^\nu\quad(c^{ij}_0=0)
  \text{ converge for }0<|s|\ll 1.
\end{equation}
We assume $c^{ij}_0=0$ without loss of generality and
expand \eqref{eq:an} into the power series. 
Then the terms $(t_i\cdots t_{j-1})^p(t_j\cdots t_{k-1})^q$ with $p\ne0$, 
$q\ne 0$, $p\ne q$ and $i<j<k$ appear only in $U_{ijk}$ and therefore if 
$p\ne0$, $q\ne 0$ and $p\ne q$, we have
\[
  c^{jk}_q pc^{ij}_p + c^{jk}_{q-p}pc^{ik}_p
  -c^{ik}_q(p-q)c^{ij}_{p-q}+c^{ik}_p(q-p)c^{jk}_{q-p}
  -c^{ij}_{p-q}qc^{ik}_q - c^{ij}_pqc^{jk}_q=0
\]
and hence
\[
 (p-q)c^{ij}_pc^{jk}_q - pc^{ij}_{p-q}c^{ik}_q + qc^{jk}_{q-p}c^{ik}_p = 0.
\]
Denoting
\begin{equation}\label{eq:anU}
 U_{ij}(t) = \sum_{\nu\in\mathbb Z\setminus\{0\}}C^{ij}_\nu t^\nu\quad
 \text{with } c^{ij}_\nu = \nu C^{ij}_\nu,
\end{equation}
we have
\begin{gather}
 u_{ij}(t) = tU'_{ij}(t),\\
 pq(p-q)\bigl(C^{ij}_pC^{jk}_q - C^{ij}_{p-q}C^{ik}_q - C^{jk}_{q-p}C^{ik}_p\bigr) 
 = 0.\label{eq:ani}
\end{gather}
Then \eqref{eq:an} is equivalent to
\begin{equation}\label{eq:a2f}
 \bigl(U_{ij}(s) + U_{jk}(t) - U_{ik}(st)\bigr)^2 = 
 V_{ij}^{ijk}(s) + V_{jk}^{ijk}(t) - V_{ik}^{ijk}(st)
\end{equation}
with suitable functions $V_{ij}^{ijk}$, $V_{jk}^{ijk}$ and $V_{ik}^{ijk}$
for $1\le i<j<k\le n$.

\begin{rem}\label{rem:a2}
If $\bigl(U_{ij}(t), U_{jk}(t), U_{ik}(t)\bigr)$ satisfies \eqref{eq:a2f} 
with suitable $V_{ij}$, $V_{jk}$ and $V_{ij}$, then
$\bigl(U_{jk}(t), U_{ij}(t), U_{ik}(t)\bigr)$ and
$\bigl(cU_{ij}(at^r), cU_{jk}(bt^r), cU_{ik}(abt^r)\bigr)$ have the same 
property for any complex numbers $a$, $b$ and $c$ and a positive integer $r$ 
with $ab\ne 0$.
\end{rem}

\begin{prop}\label{prop:A}
The solution $(U_{ij}, U_{jk}, U_{ik})$ of \eqref{eq:a2f} with \eqref{eq:anU}
is one of the followings and it satisfies
$U_{ijk}=0$.

{\rm i)}
Two of $\{U_{ij}, U_{jk}, U_{ik}\}$ are zero and the other one is any function.

{\rm ii)}
$(U_{ij}, U_{jk}, U_{ik}) = (at^r, bt^r, ct^{-r})$ for any
$a$, $b$ and $c\in\mathbb C$ and $r\in\mathbb Z\setminus\{0\}$.

{\rm iii)}
$(U_{ij}, U_{jk}, U_{ik}) = 
\bigl(\dfrac{act^r}{1-at^r}, \dfrac{bct^r}{1-bt^r}, \dfrac{abct^r}{1-abt^r}\bigr)$
with any non-zero complex numbers $a$, $b$ and $c$ and a positive integer $r$.
\end{prop}
\begin{proof}
All the solutions of the equation \eqref{eq:a2f} are obtained by 
\cite{BP} and \cite{BB} (cf.~\cite[Remark~2.3]{OO}), which implies this proposition.
But we will give a simple proof under the assumption that
the origin is at most a pole of $U_{ij}$, $U_{jk}$ and $U_{ik}$.

Suppose one of $U_{ij}, U_{jk}, U_{ik}$ is zero 
and the other two are not zero. 
If $U_{ij}=0$ and $C^{jk}_r\ne 0$, 
then we have $p(m+p)mC^{jk}_mC^{ik}_p=0$ and 
therefore $C^{ik}_p=0$ for $p\ne -r$, $C^{ik}_{-r}\ne 0$ and $C^{jk}_m=0$ for $m\ne r$.
Thus we have {\rm ii)}.  
We similarly have {\rm ii)} in the other two cases.

Hence we may assume that any one of $\{U_{ij}, U_{jk}, U_{ik}\}$ is not zero.
Define $I_{\ell m}\in\mathbb Z\setminus\{0\}$ such that $C^{\ell m}_{I_{\ell m}}\ne0$
and $C^{\ell m}_\nu = 0$ for $\nu< I_{\ell m}$.  Then \eqref{eq:ani} shows
\begin{equation}\label{eq:a2C}
 I_{ij}I_{jk}(I_{ij}-I_{jk})(C^{ij}_{I_{ij}}C^{jk}_{I_{jk}}
 - C^{ij}_{I_{ij}-I_{jk}}C^{ik}_{I_{jk}} - C^{jk}_{I_{jk}-I_{ij}}C^{ik}_{I_{ik}}) = 0.
\end{equation}

Suppose $I_{ij}>0$ and $I_{jk}>0$.  Then \eqref{eq:a2C} means $I_{ij}=I_{jk}$, which
we put $r$, and therefore \eqref{eq:ani} with $q=r$ and that with $p=q+r$ mean
\begin{align}
  pr(p-r)(C^{ij}_pC^{jk}_r - C^{ij}_{p-r}C^{ik}_r) &= 0
    \quad\text{for }p>0,\label{eq:a2sh}\\
  (q+r)qr(C^{ij}_{q+r}C^{jk}_q - C^{ij}_rC^{ik}_q) &= 0,\label{eq:a2she}
\end{align}
respectively.

If $C^{ik}_r=0$, it follows from \eqref{eq:a2sh} that $C^{ij}_p = 0$ for $p\ne r$
by the induction on $p$ and we have similarly $C^{jk}_q = 0$ for $q\ne r$ 
by the symmetry between $U^{ij}$ and $U^{jk}$ 
and finally $C^{ik}_q = 0$ for $q\ne -r$ by \eqref{eq:a2she}.
Hence this case is reduced to ii) with $r>0$.

Suppose $C^{ik}_r\ne 0$.  Then by Remark~\ref{rem:a2} we may assume 
$C^{ij}_r=C^{jk}_r=C^{ik}_r$ by a suitable transformation $(s,t)\mapsto(at,bt)$
and moreover by \eqref{eq:a2sh} that $U_{ij}=c\sum_{\nu=1}^\infty t^{r\nu}$
and similarly $U_{jk}=c'\sum_{\nu=1}^\infty t^{r\nu}$.
Then \eqref{eq:a2she} means $U_{ik} = U_{jk} + c''t^{-r}$.
Finally we have $c''=0$ by \eqref{eq:a2C} with $p=2r$ and $q=-r$
and get $U_{ij}=U_{jk}=U_{ik}$. 

Lastly we may assume $I_{ij}<0$ by Remark~\ref{rem:a2}.
Then \eqref{eq:ani} with $p=I_{ij}+I_{ik}$ and $q=I_{ik}$ implies $I_{ik}>0$ and
that with $p=I_{ij}$ and $q>0$ means $C^{jk}_q=0$ for $q\ge 0$.
Hence $I_{jk}<0$ and similarly we have $C^{ij}_p=0$ for $p\ge 0$.
Moreover \eqref{eq:ani} with $p=q+I_{ij}$ shows $C^{ik}_q=0$ for sufficiently 
large integer $q$.
Then $\bigl(U_{ij}(t^{-1}),U_{jk}(t^{-1}), U_{ik}(t^{-1})\bigr)$ 
is also a solution 
of \eqref{eq:a2f} and this case is reduced to the case when $I_{ij}>0$ and 
$I_{jk}>0$ and therefore we have ii) with $r<0$.

Note that it is easy to see that the given functions in the proposition 
satisfy $U_{ijk}=0$ (cf.~Remark~\ref{rem:a2}).
\end{proof}
\begin{rem}  If $t=e^{-x}$, then
\begin{align*}
 t\frac d{dt}\bigl(at^r\bigr) &= art^r=are^{-rx},\\
 t\frac d{dt}\Bigl(\frac{at^r}{1-at^r}\Bigr) &=
 \frac{art^r}{(1-at^r)^2} = r\sh^{-2}\frac{rx-\log a}2.
\end{align*}
\end{rem}

\section{Type $B_2$}\label{sec:B2}
In this section we study the following commuting differential operators.
\begin{equation}\label{eq:b2-0}
\begin{cases}
 P=\dfrac{\p ^2}{\p x^2}+\dfrac{\p ^2}{\p y^2} + R(x,y),\\
 Q=\dfrac{\p^4}{\p x^2\p y^2} + S \quad\text{ with }\ord S<4,\\
 [P,Q]=0.
\end{cases}
\end{equation}
Note that $P_2=P^2-2Q$ in \eqref{eq:Bn}.
First we review the arguments given in \cite{OO} and \cite{Oc}.
Since $P$ is self-adjoint, we may assume $Q$ is also self-adjoint by replacing
$Q$ by its self-adjoint part if necessary.
Here for $A=\sum a_{ij}(x,y)\frac{\p^{i+j}}{\p x^i\p y^j}$ we define
${}^t\!A=\sum (-1)^{i+j}\frac{\p^{i+j}}{\p x^i\p y^j}a_{ij}(x,y)$
and $A$ is called self-adjoint if ${}^t\!A=A$.
Then
\begin{equation}\label{eq:b2-1}
\begin{split}
  R(x,y) &= u^+(x+y)+u^-(x-y)+v(x)+w(y),\\
  Q&=\left(\frac{\p ^2}{\p x\p y} + \frac{u^+(x+y) - u^-(x-y)}2\right)^2 + 
  w(y)\frac{\p^2}{\p x^2} + v(x)\frac{\p ^2}{\p y^2}\\
   &\quad
   + v(x)w(y) + T(x,y),\\
\end{split}
\end{equation}
and the function $T(x,y)$ satisfies
\begin{equation}\label{eq:b2-2}
\begin{split}
  2\frac{\p T(x,y)}{\p x} &= \bigl(u^+(x+y)-u^-(x-y)\bigr)\frac{\p w(y)}{\p y}
    + 2w(y)\frac{\p}{\p y}\bigl(u^+(x+y)-u^-(x-y)\bigr),\\
  2\frac{\p T(x,y)}{\p y} &= \bigl(u^+(x+y)-u^-(x-y)\bigr)\frac{\p v(x)}{\p x}
    + 2v(x)\frac\p{\p x}\bigl(u^+(x+y)-u^-(x-y)\bigr).
\end{split}
\end{equation}
On the other hand, if a function $T(x,y)$ satisfies \eqref{eq:b2-2}
for suitable functions $u^\pm(t)$, $v(t)$ and $w(t)$, then \eqref{eq:b2-0} is
valid for $R(x,y)$ and $Q$ defined by \eqref{eq:b2-1}.

We have the compatibility condition 
\begin{equation}\label{eq:b2comp}
\begin{aligned}
\frac{\p}{\p x}\Bigl(&\bigl(u^+(x+y)-u^-(x-y)\bigr)\frac{\p v(x)}{\p x}
  + 2v(x)\frac\p{\p x}\bigl(u^+(x+y)-u^-(x-y)\bigr)\Bigr)\\
=&
\frac\p{\p y}\Bigl(\bigl(u^+(x+y)-u^-(x-y)\bigr)\frac{\p w(y)}{\p y}
  + 2w(y)\frac\p{\p y}\bigl(u^+(x+y)-u^-(x-y)\bigr)\Bigr)
\end{aligned}
\end{equation}
for the existence of $T(x,y)$.

\begin{defn}[Duality in $B_2$]\label{defn:dual}
Under the coordinate transformation
\begin{equation}
 (x,y)\mapsto\left(\frac{x+y}{\sqrt 2},\frac{x-y}{\sqrt 2}\right)
\end{equation}
the pair $(P,\frac14 P^2 - Q)$ also satisfies \eqref{eq:b2-0},
which we call the {\it duality\/} of the commuting differential operators of 
type $B_2$.
\end{defn}

Denoting $\p_x=\frac\p{\p x}$, $\p_y=\frac\p{\p y}$ and put 
\[
 L=P^2 - 4Q - (\p_x^2-\p_y^2 +v(x)-w(y))^2 - 2u^-(x-y)(\p_x+\p_y)^2 - 2u^+(x+y)(\p_x-\p_y)^2.
\]
Then the order of $L$ is at most 2 and the second order term of $L$ equals
\begin{multline*}
  2(u^++u^-+v+w)(\p_x^2+\p_y^2) - 4(u^+-u^-)\p_x\p_y - 4w\p_x^2 - 4v\p_y^2\\
 - 2(v-w)(\p_x^2-\p_y^2) -2u^-(\p_x+\p_y)^2-2u^+(\p_x-\p_y)^2 = 0.
\end{multline*}
Since $L$ is self-adjoint, $L$ is of order at most 0 and the 0-th order term of $L$
equals
\begin{multline*}
 (\p_x^2+\p_y^2)(u^++u^-+v+w) + (u^++u^-+v+w)^2 - 4(vw+T) - 2\p_x\p_y(u^+-u^-)\\
 -(\p_x^2-\p_y^2)(v-w) = (u^++u^-+v+w)^2 - 4(vw+T)
\end{multline*}
and therefore we have the following proposition.
%%%%%%%%%%
\begin{prop} {\rm i)}
By the duality in Definition~\ref{defn:dual} the pair $\bigl(R(x,y),T(x,y)\bigr)$ changes into
$\bigl(\tilde R(x,y),\tilde T(x,y)\bigr)$ with
\begin{align}
 \begin{cases}
   \tilde R(x,y) = v\Bigr(\dfrac{x+y}{\sqrt2}\Bigl) + w\Bigl(\dfrac{x-y}{\sqrt2}\Bigr) 
                + u^+\bigl(\sqrt{2}x\bigr) + u^-\bigl(\sqrt{2}y\bigr),\\
   \tilde T(x,y) 
             = \dfrac14\tilde R(x,y)^2
             - v\Bigl(\dfrac{x+y}{\sqrt2}\Bigr)w\Bigl(\dfrac{x-y}{\sqrt2}\Bigr)
             - T\Bigl(\dfrac{x+y}{\sqrt2},\dfrac{x-y}{\sqrt2}\Bigr).
 \end{cases}
\end{align}

{\rm ii)} Combining the duality with the scaling map $R(x,y)\mapsto c^{-2}R(cx,cy)$, 
the following pair $\bigl(R^d(x,y),T^d(x,y)\bigr)$ defines commuting differential operators 
if so is $\bigl(R(x,y),T(x,y)\bigr)$.  
This $R^d(x,y)$ is also called the {\it dual\/} of $R(x,y)$.

\begin{align}
 \begin{cases}
   R^d(x,y) =  v(x+y) + w(x-y) + u^+(2x) + u^-(2y),\\
   T^d(x,y) = \frac14 R^d(x,y)^2 - v(x+y)w(x-y) - T(x+y,x-y).
 \end{cases}
\end{align}
\end{prop}
\iffalse
The $B_2$-invariant solutions of \eqref{eq:b2comp} are determined by
\cite{OO} (cf.~\cite{OOS}).  They are
\begin{align}\label{eq:b2-para5}
 &\begin{cases}
 u^+(t)=u^-(t)=A\wp(t)+A_0,\\
 v(t)=w(t)=
  \dfrac{C_4\wp(t)^4+C_3\wp(t)^3+C_2\wp(t)^2+C_1\wp(t)+C_0}{\wp'(t)^2},\\
   T(x,y)= A\bigl(\wp(x)-\wp(y)\bigr)^{-2}
      \Bigl(C_4\wp(x)^2\wp(y)^2+\dfrac{C_3}2\wp(x)^2\wp(y)\\
       \quad\quad\quad\quad
       +\dfrac{C_3}2\wp(x)\wp(y)^2 +
       C_2\wp(x)\wp(y)+ \dfrac{C_1}2\wp(x)+\dfrac{C_1}2\wp(y) + C_0\Bigr),
 \end{cases}
\intertext{and the dual of \eqref{eq:b2-para5} and}
\label{eq:b2-para4}
&\begin{cases}
 u^{\pm}(t) = A_2\dfrac{(\wp(\frac t2)-e_3)^2}{\wp'(\frac t2)^2}
   + A_1\wp(t) + A_0,\\
 v(t)=w(t) =\dfrac{C_2\wp(t)^2 + C_1\wp(t) + C_0}{\wp(t)-e_3},\\
 T(x,y) = .
\end{cases}
\end{align}
The function $T(x,y)$ in \eqref{eq:b2-para5} is due to \cite[Lemma~7.3]{OS}.
Here $\wp(t)=\wp(t;2\omega_1,2\omega_2)$ is the Weierstrass elliptic
function with the fundamental periods $2\omega_1$ and $2\omega_2$
and we allow that the periods are infinity in the above.


If the periods of $\wp(t)$ are finite, we may rewrite \eqref{eq:b2-para5} as
\begin{align}
\allowdisplaybreaks
 &\begin{cases}\label{eq:b2el}
   u^{\pm}(t)= A\wp(t)+A_0, \\
   v(t)=w(t)=\displaystyle\sum_{\nu=0}^3C_\nu\wp(t+\omega_i) + C_4,\\
   T(x,y) = A\Bigl(\displaystyle\sum_{\nu=0}^3C_\nu\wp(x+\omega_i)\cdot\wp(y+\omega_i)
            \\%- C_4\bigl(\wp(x+y)+\wp(x-y)-1\bigr)\\
   \quad\quad\quad\quad
         -\dfrac{v(x)+w(y)}2\bigl(\wp(x+y)+\wp(x-y)\bigr)\Bigr)
 \end{cases}\\
\intertext{(cf.~\cite[Corollary~5.2]{O2}) and if $\wp(t)$ has a finite period, then}
\allowdisplaybreaks
 &\begin{cases}
   u^{\pm}(t)= A\sh^{-2}\lambda t+A_0, \\
   v(t)=w(t)=C_1\sh^{-2}\lambda t + C_2\ch^{-2}\lambda t + C_3\sh^2\lambda t\\
            \quad\quad\quad\quad
             + C_4\sh^22\lambda t + C_0,\\
   T(x,y) = A\Bigl(C_1\sh^{-2}\lambda x\cdot\sh^{-2}\lambda y -
            C_2\ch^{-2}\lambda x\cdot\ch^{-2}\lambda y\\
            \quad\quad\quad\quad
               + C_4(\sh^2\lambda x + \sh^2\lambda y + 
               2\sh^2\lambda x\cdot\sh^2\lambda y)\\
            \quad\quad\quad\quad
            -\dfrac{v(x)+w(y)}{2}\bigl(\sh^{-2}\lambda(x+y)
               +\sh^{-2}\lambda(x-y)\bigr)\Bigr)
 \end{cases}
\allowdisplaybreaks
\intertext{or}
 &\begin{cases}
   u^{\pm}(t)= At^{-2}+A_0, \\
   v(t)=w(t)=C_1t^{-2} + C_2t^2 + C_3t^4 + C_4t^6 + C_0,\\
   T(x,y) = A\Bigl(\dfrac{C_1}{x^2y^2} + C_3(x^2+y^2)+C_4(x^4+y^4+3x^2y^2)\\
             \quad\quad\quad\quad
            -\bigr(v(x)+w(y)\bigl)\dfrac{x^2+y^2}{(x^2-y^2)^2}\Bigr),
 \end{cases}
\end{align}
which are given by \cite[Proposition~6.1 and 6.3]{O2}.
Note that
\begin{align}
 \begin{split}
  &\quad C_1\sh^{-2}\lambda t + C_2\ch^{-2}\lambda x + 
  C_3\sh^2\lambda t + C_4\sh^22\lambda t + C_0\\
  &=
  (C_1+C_2)\sh^{-2}\lambda t -4C_2\sh^{-2}2\lambda t
  +\dfrac{C_3}2\ch2t + \dfrac{C_4}2\ch4t\\
  &\quad +C_0 - \dfrac{C_3}2-\dfrac{C_4}2\\
  &=C_1\sh^{-2}\lambda t + (C_3-C_2)\tanh^2\lambda t
    +(C_3+4C_4)\tanh^2\lambda t\cdot\sh^2\lambda t\\
   &\quad
    +4C_4\tanh^2\lambda t\cdot\sh^4\lambda t
    +C_0+C_2.
 \end{split}
\end{align}
Put
\begin{equation}
  S(u(t),v(t)) = \bigl(u(x+y)-u(x-y)\bigr)\p_yv(y)
   +2w\p_y\bigl(u(x+y)-u(x-y)\bigr).
\end{equation}
Then as is given in \cite[Corollary~2.2]{O2}, it is easy to have
\begin{equation}
 S(\phi(t),\phi(t)) =
 \p_x\Bigl(2\phi(x)\cdot\phi(y) - 
   \bigl(\phi(x)+\phi(y)\bigr)\bigl(\phi(x+y)+\phi(x-y)\bigr)\Bigr)
\end{equation}
for
\begin{equation}
 \phi(t)=C_1\wp(t)+C_2\quad\text{with }C_1,\ C_2\in\mathbb C
\end{equation}
and hence
\begin{multline}
 S(\wp(t),\wp(t+\omega_\nu)) = \\
 \p_x\Bigl(2\wp(x+\omega_\nu)\cdot\wp(y+\omega_\nu) - 
   \bigl(\wp(x+\omega_\nu)+\wp(y+\omega_\nu)\bigr)
   \bigl(\wp(x+y)+\wp(x-y)\bigr)\Bigr)
\end{multline}
for $\nu=0$, $1$, $2$ and $3$, which shows \eqref{eq:b2el}.
Moreover if follows from \eqref{eq:wp2} that
\begin{multline}
 S(\wp(t),\wp(2t)) = \\
   \p_x\Bigl(\frac12\sum_{\nu=0}^3\wp(x+\omega_\nu)\cdot\wp(y+\omega_\nu) - 
   \bigl(\wp(2x) + \wp(2y)\bigr)
   \bigl(\wp(x+y)+\wp(x-y)\bigr)\Bigr)
\end{multline}
and
\begin{multline}
 S(\wp(\frac t2),\wp(t)) = \\
   \p_x\Bigl(\frac12\sum_{\nu=0}^3\wp(\frac x2+\omega_\nu)
     \cdot\wp(\frac y2+\omega_\nu) - 
   \bigl(\wp(x) + \wp(y)\bigr)
   \bigl(\wp(\frac{x+y}2)+\wp(\frac{x-y}2)\bigr)\Bigr).
\end{multline}
Hence we have
\begin{multline}
 S(\wp(\frac t2+\omega_i),\wp(t)) = \\
   \p_x\Bigl(\frac12\sum_{\nu=0}^3\wp(\frac x2+\omega_i+\omega_\nu)
     \cdot\wp(\frac y2+\omega_\nu)\\
   - \bigl(\wp(x) + \wp(y)\bigr)
   \bigl(\wp(\frac{x+y}2+\omega_i)+\wp(\frac{x-y}2+\omega_i)\bigr)\Bigr)
\end{multline}
and hence
\begin{multline}
 S(4\wp(t;2\omega_1,4\omega_2),\wp(t)) = \\
   \p_x\Bigl(\frac12\sum_{i\in\{1,3\}}\sum_{\nu=0}^3\wp(\frac x2+\omega_i+\omega_\nu)
     \cdot\wp(\frac y2+\omega_\nu)\\
   - \sum_{i\in\{1,3\}}\bigl(\wp(x) + \wp(y)\bigr)
   \bigl(\wp(\frac{x+y}2+\omega_i)+\wp(\frac{x-y}2+\omega_i)\bigr)\Bigr).
\end{multline}

\begin{align*}
 \coth y\cdot\bigl(\sh(x+y)-\sh(x-y)\bigr)
  &= 2\coth y\cdot\ch x\cdot\sh y = 2\ch x\cdot\ch y\\
  &=\ch(x+y)-\ch(x-y),\\
 \coth(x-y)\cdot\bigl(e^{2x} - e^{2y}\bigr)
  &= \ch(x-y)\cdot2e^{x+y}=e^{2x}+e^{2y},\\
 \coth(x-y)\cdot\bigl(e^{4x} - e^{4y}\bigl)
  &= \bigl(e^{2x}+e^{2y}\bigr)^2 = e^{4x}+2e^{2(x+y)}+e^{4y}.
\end{align*}

Similarly we have
\begin{align}
 &\begin{cases}
   u^{\pm}(t)= A_1\wp(t;2\omega_1,2\omega_2)+
          A_2\wp(t+2\omega_1;4\omega_2,2\omega_2)+A_0, \\
   v(t)=w(t)=C_1\wp(t;2\omega_1,2\omega_2) + C_2\wp(t+\omega_2;2\omega_1,2\omega_2)
         + C_0\\
   T(x,y) = ???\\
    \quad\quad\quad\quad + 
    ???,
 \end{cases}
\allowdisplaybreaks\\
 &\begin{cases}
   u^{\pm}(t)= A_1\sh^{-2}\lambda t+A_2\ch 2\lambda t+A_0, \\
   v(t)=w(t)=C_1\sh^{-2}\lambda t + C_2\ch^{-2}\lambda t + C_0,\\
   T(x,y) = ???\\
    \quad\quad\quad\quad + 
    ???,
 \end{cases}
\allowdisplaybreaks\\
 &\begin{cases}
   u^{\pm}+(t)= A_1t^{-2}+A_2t^2+A_0, \\
   v(t)=w(t)=C_1t^{-2} + C_2t^2 + C_0,\\
   T(x,y) = \dfrac{2A_1(2C_1+2C_2x^2y^2+C_0x^2+C_0y^2)}{(x^2-y^2)^2}\\
    \quad\quad\quad\quad + 
    2A_2(2C_2x^2y^2+C_0x^2+C_0y^2),
 \end{cases}
\end{align}
\fi
Now we give a list of the solutions of \eqref{eq:b2comp} and \eqref{eq:b2-2}.
They are suitable limits of the invariant solutions studied in \cite{OO}
and many of them are given in \cite{Oc}.

Case I: (Any-$A_1$)$+$(Any-$A_1$)\quad 
    $v=w=0$ and $u$ and $v$ are arbitrary functions.

Case II: $u^+=u^-$, $v=w$ and $(u^+;v)$ is in the following list.
\begin{gather*}
  (\langle \sh^{-2}\lambda t\rangle;\ \langle \sh^{-2}2\lambda t,\ 
   \sh^{-2}\lambda t,\ \ch2\lambda t,\ \ch4\lambda t\rangle),\label{eq:list-top}
   \tag{\rm Trig-$B_2$}\\ 
  (\langle \sh^{-2}\lambda t,\ \sh^{-2}2\lambda\rangle;\ 
   \langle\sh^{-2}2\lambda t,\ \ch 4\lambda t\rangle)
  \tag{\rm Trig-$B_2$-S}.
\end{gather*} 

Case III: $u^+=u^-$, $(u^+;v,w)$ is in the following list.
\begin{gather*}
  (\langle \ch2\lambda t\rangle;\ 
   \langle \sh^{-2}\lambda t,\ \sh^{-2}{2\lambda t}\rangle,\  
   \langle \sh^{-2}\lambda t,\ \sh^{-2}{2\lambda t}\rangle)
  \tag{\rm Toda-$D_2^{(1)}$-bry},\\
  (\langle \ch\lambda t,\ \ch2\lambda t\rangle;\ 
   \langle \sh^{-2}\lambda t\rangle,\  \langle \sh^{-2}\lambda t\rangle)
  \tag{\rm Toda-$D_2^{(1)}$-S-bry},\\
  (\langle e^{-2\lambda t}\rangle;\ 
   \langle e^{2\lambda t},\ e^{4\lambda t}\rangle,\ 
   \langle \sh^{-2}\lambda t,\ \sh^{-2}{2\lambda t}\rangle)
  \tag{\rm Toda-$B_2^{(1)}$-bry},\\
  (\langle e^{-\lambda t},\ e^{-2\lambda t}\rangle;\  
   \langle e^{2\lambda t}\rangle,\ \langle \sh^{-2}\lambda t\rangle)
  \tag{\rm Toda-$B_2^{(1)}$-S-bry}.
\end{gather*}

Case IV: $v=w$, $(u^+,u^-;v)$ is in the following list.
\begin{gather*}
  (0,\ \langle\sh^{-2}\lambda t\rangle;\ 
  \langle e^{-2\lambda t}, e^{-4\lambda t}, e^{2\lambda t}, e^{4\lambda t}\rangle)
  \tag{\rm Trig-$A_1$-bry},\\
  (0,\ \langle \sh^{-2}\lambda t,\ \sh^{-2}{2\lambda t}\rangle;\ 
   \langle e^{-4\lambda t},\ e^{4\lambda t}\rangle)
  \tag{\rm Trig-$A_1$-S-bry}.
\end{gather*}

Case V: $(u^+,u^-,v,w)$ is in the following list.
\begin{gather*}
  (0,\ \langle e^{-\lambda t}\rangle,\ 
   \langle e^{\lambda t},\ e^{2\lambda t}\rangle,\ 
   \langle e^{-\lambda t},\ e^{-2\lambda t}\rangle)
  \tag{\rm Toda-$C_2^{(1)}$},\\
  (0,\ \langle e^{-\lambda t},\ e^{-2\lambda t}\rangle,\ 
   \langle e^{2\lambda t}\rangle,\ 
   \langle e^{-2\lambda t}\rangle)
   \tag{\rm Toda-$C_2^{(1)}$-S}.
   \label{eq:list-last}
\end{gather*}
In the above $\langle\ \rangle$ means an arbitrary linear combination of 
given functions and, for example, (Trig-$B_2$) implies
\[
 \begin{cases}
 u^+(t)=u^-(t)=C_0\sh^{-2}\lambda t,\\ 
 v(t)=w(t)=C_1\sh^{-2}2\lambda t+C_2\sh^{-2}\lambda t
  +C_3\ch2\lambda t+C_4\ch4\lambda t
 \end{cases}
\]
with any complex numbers $C_0,C_1,\dots,C_4$ and 
a suitable $\lambda\in\mathbb C\setminus\{0\}$.

According to our assumption, put
\begin{equation}\label{eq:series}
\begin{gathered}
  t = e^{-y},\quad s = e^{-x+y},\\
  u^+(x+y) = \sum_{i\ge r}u^+_i s^it^{2i},\ 
  u^-(x-y) = \sum_{i\ge r}u^-_is^i,\\
  v(x) = \sum_{j\ge r'}v_j s^jt^j,\ %\quad\text{and}\quad
  w(y) = \sum_{j\ge r''}w_jt^j,\\
  u^{\pm}_i=v_j=w_k=0\quad\text{if } i<r,\, j<r' \text{ and }k<r''.
\end{gathered}
\end{equation}
\begin{equation}\label{eq:exp1}
\begin{aligned}
  {}&\sum_{\substack{ i\ge r\\j\ge r'}}
       (i+j)(2i+j)v_j(u_i^+t^{2i+j} - u_i^-t^{j})s^{i+j}\\
  {}&= \sum_{\substack{i\ge r\\j\ge r''}}
       \Big((2i+j)(i+j)w_ju_i^+t^{2i+j}-(2i-j)(i-j)w_ju_i^-t^{j}\Big)s^i
\end{aligned}
\end{equation}
and the coefficients of $s^pt^q$ mean
\begin{equation}\label{eq:b2i0}
   pqv_{2p-q}u^+_{q-p} - p(2p-q)v_qu^-_{p-q}
 = q(q-p)w_{q-2p}u^+_p - (2p-q)(p-q)w_qu^-_p.
\end{equation}
Putting
\begin{equation}\label{b2:series2}
  \begin{cases}
    U^\pm(t) = \sum_{i\ge r}U^\pm_it^i,\ 
    V(t) = \sum_{j\ge r'}V_jt^j \text{ and }
    W(t) = \sum_{k\ge r''}W_kt^k,\\
    u^\pm(t)=t(U^\pm)'(t)+u^\pm_0,\ v(t)=tV'(t)+v_0
    \text{ and }w(t)=tW'(t)+w_0,
  \end{cases}
\end{equation}
we have
\begin{equation}\label{eq:b2i}
  pq(2p-q)(p-q)\bigl(V_{2p-q}U^+_{q-p} + V_qU^-_{p-q}
   + W_{q-2p}U^+_p - W_qU^-_p\bigr) = 0,
\end{equation}
which is equivalent to
\begin{multline}\label{eq:b2fun}
 V(st)\bigl(U^+(st^2) + U^-(s)\bigr) + 
 W(t)\bigl(U^+(st^2) - U^-(s)\bigr)\\
 = F_1(st^2) + F_2(s) + G_1(st) + G_2(t)
\end{multline}
with suitable functions $F_1$, $F_2$, $G_1$ and $G_2$ 
(cf.~\cite[Proposition~2.4]{Oc}).
Thus we have the following proposition.
\begin{prop}
For the functions $(U^{\pm}, V, W, F_1,F_2,G_1,G_2)$ satisfying \eqref{eq:b2fun}
we have the commuting differential operators \eqref{eq:b2-0} and \eqref{eq:b2-1} 
by putting
\begin{equation}
 \begin{cases}
   u^{\pm}(t) =\p_tU^{\pm}(e^t)+C',\ v(t)=\p_tV(e^t)+C,\ w(t)=\p_tW(e^t)+C,\\
   T(x,y) = \dfrac12\Bigl(\p_x^2-\p_y^2\Bigr)
            \Bigl(V(e^x)\bigl(U^+(e^{x+y})+U^-(e^{x-y})\bigr) - G_1(e^x)\Bigr)\\
            \quad\quad\quad\quad
            +C\bigl(u^+(x+y)+u^-(x-y)\bigr),\\
  \quad\quad C,\ C'\in\mathbb C.
 \end{cases}
\end{equation}
\end{prop}

Now we put
\begin{equation}
\begin{aligned}
\mathcal S(B_2)&=\bigl\{\bigl(U^+(t),U^-(t),V(t), W(t)\bigr);\,U^\pm, V \text{ and }W 
  \text{ are meromorphic}\\
 &\text{in a neighborhood of  0 and they satisfy }\eqref{eq:b2fun}\bigr\}.
\end{aligned}
\end{equation}

\begin{rem}\label{rem:b2-0}
 i)
Since the constant terms $U^{\pm}_0$, $V_0$ and $W_0$ have no effect on 
the equation \eqref{eq:b2i} and on the original functions 
$u^\pm$, $v$ and $w$, we will identify two functions appeared in the solutions of 
\eqref{eq:b2i} if they only differ in their constant terms.

ii) If $\bigl(U^+(t),U^-(t)\bigr)=0$ or $\bigl(W(t),V(t)\bigr)=0$, then \eqref{eq:b2i} is always true.
We call such $(U^+,U^-,W,V)\in \mathcal S(B_2)$ a {\it trivial solution\/} of 
\eqref{eq:b2fun}.
\end{rem}

We summarize elementary transformations acting on $\mathcal S(B_2)$.

\begin{lem}\label{rem:b2}
Let $\bigl(U^+(t),U^-(t),V(t),W(t)\bigr) \in\mathcal S(B_2)$.

{\rm i) (dual)}
$\bigl(V(t), W(t), U^+(t^2), U^-(t^2)\bigr)\in\mathcal S(B_2)$.

{\rm ii) (bilinear)}
If $\bigl(U^+(t),U^-(t),S(t), T(t)\bigr)\in\mathcal S(B_2)$, then
$\bigl(aU^+(t), aU^-(t), bV(t)+cS(t), bW(t)+cT(t)\bigr)\in\mathcal S(B_2)$ for 
$a,b,c\in\mathbb C$.

{\rm iii) (translations)}
$\bigl(U^+(ab^2t), U^-(bt), V(abt), W(at)\bigr)\in\mathcal S(B_2)$ for 
$a,b\in\mathbb C\setminus\{0\}$.

{\rm iv) (scaling)}
If $\bigl(U^+(t^r),U^-(t^r),W(t^r),V(t^r)\bigr)$ is well-defined for a suitable $r\in\mathbb Q\setminus\{0\}$, it is in $\mathcal S(B_2)$.

{\rm v) (symmetry)}
If $W(t)$ is a rational function, the reflection $(x,y)\mapsto(x,-y)$ can 
be applied to the solution and then 
$\bigl(U^-(t),U^+(t),V(t),-W(t^{-1})\bigr)\in\mathcal S(B_2)$.

{\rm vi) (symmetry)}
If $U^-(t)$ is a rational function, the reflection $(x,y)\mapsto(y,x)$ can 
be applied to the solution and then 
$\bigl(U^+(t),-U^-(t^{-1}),V(t),W(t)\bigr)\in\mathcal S(B_2)$.
\end{lem}
%\begin{proof}
The lemma is a direct consequence of the definition of $\mathcal S(B_2)$.
For example, i) follows from
\begin{multline*}
 U^+(t^2s^2)\bigl(V(ts^2)+W(t)\bigr)+U^-(s^2)\bigl(V(ts^2)-W(t)\bigr)\\
 =  F_2(t) + F_1(s^2) + G_2(t^2s^2) + G_1(ts^2).
\end{multline*}
%\end{proof}

Note that the transformation in Lemma~\ref{rem:b2}~vi) is equals to a certain
composition of transformations in Lemma~\ref{rem:b2}~i), iv) and v).

\begin{defn}
If a solution of \eqref{eq:b2fun} obtained by applying transformations in 
Lemma~\ref{rem:b2} to an original solution, it is called a 
{\it standard transformation\/} of the original solution.
\end{defn}

We will study non-trivial solutions of \eqref{eq:b2fun}.
Considering standard transformations, we may assume
\begin{equation}
  (U^+_r,U^-_r) = (1,1) \text{ or } (1,0) \text{ or } (0,1).
\end{equation}

\begin{prop}\label{prop:rat}
Suppose $\bigl(U^+(t),U^-(t),V(t),W(t)\bigr)$ is a non-trivial solution of \eqref{eq:b2fun} with \eqref{b2:series2}.

{\rm i)} $U^\pm(t)$, $V(t)$ and $W(t)$ are rational functions.

{\rm ii)} {\rm (}\cite[Theorem~2.3]{Oc}{\rm )}
If $W(t)$ has a pole at $t=1$, then $U^+(t)=U^-(t)$
and $W(t^{-1})+W(t)=0$.
If $U^-(t)$ has a pole at $t=1$, then $V(t)=W(t)$ and $U^-(t^{-1})+U(t)=0$.

Here we note that this equality is interpreted in the sense of 
Remark~\ref{rem:b2-0}.

{\rm iii)} {\rm (}\cite[Corollary~3.8]{Oc}{\rm )} If at least two of
$\{U^+(t),U^-(t),V(t),W(t)\}$ have poles in $\mathbb C\setminus\{0\}$,
$(U^+,U^-,V,W)$ is a standard transformation of a solution given in the list
\eqref{eq:list-top} -- {\rm (Toda-$D_2^{(1)}$-S-bry)}.
\end{prop}
\begin{proof} i)
The equation \eqref{eq:b2i} shows $W_{q-2r}U^+_r=W_qU^-_r$ if $q>2|r|+|r'|$.
Hence $W(t)$ is a rational function and therefore so are $U^-(t)$, $U^+(t)$
and $V(t)$ because of Lemma~\ref{rem:b2}~i) and v).
\end{proof}

\begin{lem}\label{lem:b2-1}
{\rm i)} If $V(t)$ has a pole at the origin, then $U^+(t)$ and $U^-(t)$ are holomorphic
at the origin.

{\rm ii)} If $U^+(t)$ has a pole at the origin, then $V(t)$ and $W(t)$ are holomorphic
at the origin.
\end{lem}
\begin{proof}
If $r<0$ and $r'<0$ with $V_{r'}\ne0$, the coefficients of $s^{r+r'}t^{r'}$ and that of $s^{r+r'}t^{2r+r'}$ in \eqref{eq:b2fun} show $V_{r'}U^-_r=V_{r'}U^+_r=0$, 
which contradicts to $(U^+_r,U^-_r)\ne 0$.
Thus we have i) and then ii) by Lemma~\ref{rem:b2}~i).
\end{proof}

\begin{thm}\label{thm:B2}
Any non-trivial solution of \eqref{eq:b2fun} corresponds to a standard transformation
of a solution in the list \eqref{eq:list-top} -- \eqref{eq:list-last}.
\end{thm}
\begin{proof}
We will prove this theorem divided into several cases.

{\bf Case 1}: One of $U^+, U^-,V, W$ is zero.\newline
Proposition~\ref{prop:rat} assures that we may suppose $V=0$.
Then \eqref{eq:b2i} turns into
\begin{equation}\label{eq:W0}
 pq(2p-q)(p-q)(W_{q-2p}U^+_p - W_qU^-_p)=0.
\end{equation}

{\bf Case 1-1}: $V=0$, $W_{r''}\ne0$ and $(U^+_r,U^-_r)=(1,1)$.\newline
Suppose $\bar r:=-r>0$.
Then \eqref{eq:W0} with $p=-\bar r$ and $q=r''-2\bar r$
shows \[\bar r(r''-2\bar r)r''(r''-\bar r)W_{r''}U^+_r=0\]
and hence $r''=\bar r$ or $2\bar r$.
Since $\bar rq(2\bar r+q)(\bar r+q)(W_{q+2\bar r}U^+_r-W_qU^-_r)=0$,
\begin{equation}\label{eq:V0W}
 W(t)=at^{\bar r}(1-t^{\bar r})^{-1} + bt^{2{\bar r}}(1-t^{2\bar r})^{-1}
\end{equation}
Since $W(t)=at^{\bar r}(1+t^{\bar r})^{-1}$ if $2a+b=0$, we may assume
$W(t)$ has a pole at $t=1$ by applying a transformation in Lemma~\ref{rem:b2}~iii)
and hence $U^+(t)=U^-(t)$ by  Proposition~\ref{prop:rat}~ii).

On the other hand, \eqref{eq:W0} with $q=r''$ and that with $q=2p+r''$ show
\[
\begin{cases}
 pr''(2p-r'')(p-r'')W_{r''}U^-_p=0&\text{ for }p>0,\\
 p(2p+r'')r''(p+r'')W_{r''}U^+_p=0&\text{ for }p<0.
\end{cases}
\]
Thus we can conclude
\begin{equation}\label{eq:W0U}
 U^+(t)=U^-(t)=ct^{-\bar r}+dt^{-\frac{\bar r}2}+
 et^{\bar r}+ft^{\frac{\bar r}2}\quad\text{ with }bd=bf=0
\end{equation}
because $\bigl(U^+(t),U^-(t),0,bt^{2{\bar r}}(1-t^{2\bar r})^{-1}\bigr)\in\mathcal S(B_2)$.

If $r>0$, $rr''(2r-r'')(r-r'')W_{r''}U^-_r=0$ and
therefore $r''=2r$ or $r''=r$. Then by putting $\bar r=r$, 
the equation \eqref{eq:W0} with $p=r$ and $q> r''$ implies
\eqref{eq:V0W} and hence the same argument as above 
proves \eqref{eq:W0U}.

Hence the solution corresponds to a standard transformation of Case IV.

{\bf Case 1-2}: $V=0$ and $(U^+_r,U^-_r)=(1,0)$ or $(0,1)$.\newline
We have $q(2r-q)(r-q)W_q=0$ or $pq(2r-q)(r-q)W_{q-2r}=0$.
Hence $W(t)=at^{\bar r}+bt^{2\bar r}$ with $b\ne0$ and 
$\bar r\in\mathbb Z\setminus\{0\}$

If $a=0$, then \eqref{eq:W0} with $q=2\bar r$ implies
$U^-_p=0$ for $p\ne0$, $\bar r$, $2\bar r$.
If $a\ne0$, then \eqref{eq:W0} with $(p,q)=(\frac{\bar r}2,3\bar r)$ implies
$U^+_{\frac{\bar r}2}=0$, that with $q=2\bar r$
implies $U^-_p=0$ for $p\ne0$, $\bar r$, $2\bar r$
and that with $q=\bar r$ implies $U^-_p=0$ for $p\ne0$, 
$\pm\frac{\bar r}2$, $\bar r$.
Hence $U^-(t)=c^-t^{\bar r}+d^-t^{2\bar r}$ with $ad^-=0$.

Since $\bigl(U^+(t^{-1}),U^-(t^{-1}),0,W(t^{-1})\bigr)\in\mathcal S(B_2)$,
we have $U^+(t)=c^+t^{-\bar r}+d^+t^{-2\bar r}$ with $ad^+=0$
and the solution corresponds to the standard transform of Case V.

Now we may assume that none of $U^\pm(t),V(t),W(t)$ is zero and
\begin{equation}
 V_{r'}\ne0\quad\text{and}\quad W_{r''}\ne0.
\end{equation}
Lemma~\ref{lem:b2-1} and Lemma~\ref{rem:b2}~i) assure that we may assume 
$r>0$ except for the following case.

{\bf Case 2}: $W$ and $U^-$ have poles and $V$ and $U^+$ are holomorphic at the origin.\newline
Then $r''<0$ and $r<0$.
Put $\bar r=-r$.
The coefficients of $s^{-\bar r}t^q$ in \eqref{eq:b2fun} 
imply $W(t) = at^{-2\bar r} + bt^{-\bar r}$.

{\bf Case 2-1}: $W(t)=t^{-2\bar r} + bt^{-\bar r}$.\newline
The coefficients of $s^pt^{-2\bar r}$ imply $U^-(s)=s^{-\bar r}$.
Note that $U^-(t^{-1})$ and $W(t^{-1})$ are holomorphic at the origin and
$\bigl(U^+(t^{-1}), U^-(t^{-1}), V(t^{-1}), W(t^{-1})\bigr)\in\mathcal S(B_2)$.
If $U^+(t^{-1})$ has a pole at the origin, Lemma~\ref{lem:b2-1} assures that 
$V(t^{-1})$ is holomorphic there.
Hence we may assume $r>0$ by using a transformation in Lemma~\ref{rem:b2}~i) 
if necessary.

{\bf Case 2-2}: $W(t) = t^{-\bar r}$.\newline
The coefficients $s^pt^{-\bar r}$ in \eqref{eq:b2fun} imply
$U^-(s)=s^{-\bar r}+cs^{-\frac{\bar r}2}$
and this case is reduced to the previous case by Lemma~\ref{rem:b2}~i).

Now we may assume 
\begin{equation}
  r>0.
\end{equation}

{\bf Case 3}: $r>0$ and $r'>0$.\newline
Putting $p=r$ in \eqref{eq:b2i}, we have
\begin{equation}\label{eq:b2top}
  q(2r-q)(r-q)(W_{q-2r}U^+_r - W_qU^-_r)= 0.
\end{equation}

{\bf Case 3-1}: $(U^+_r,U^-_r)=(1,0)$ or $(0,1)$.\newline
Owing to Lemma~\ref{rem:b2}~v), we may assume $U^-_r=0$.
Then \eqref{eq:b2top} with $q=r''+2r$ means $r''=-2r$ or $r''=-r$
and \eqref{eq:b2i} with $q=r''$ means $U^-=0$.
Hence this case is reduced to Case 1.

{\bf Case 3-2}: $(U^+_r,U^-_r)=(1,1)$.\newline
The equation \eqref{eq:b2top} with $q=r''$ means $r''=r$ or $r''=2r$.

Note that \eqref{eq:b2top} means $W(t)=at^r(1-t^r)^{-1} + bt^{2r}(1-t^{2r})^{-1}$.
Since $U^\pm$, $V$ and $W$ are holomorphic at the origin, 
Lemma~\ref{rem:b2}~i) assures that if $r'\ne r''$,
this case is reduced to Case 3-1.
Hence we may assume $r'=r''$ and therefore $(V_{r'},W_{r'})=(1,1)$
by a suitable translation $s\mapsto as$.
It also follows from Lemma~\ref{rem:b2}~i) that
$U^-(s)=cs^{\frac {r'}2}(1-s^{\frac {r'}2})^{-1} + ds^{r'}(1-s^{r'})^{-1}$
and then this case is reduced to Proposition~\ref{prop:rat}~iii).

{\bf Case 4}: $r>0$ and $r'<0$.\newline
Using the transformation in Lemma~\ref{rem:b2}~v) if necessary, we
may assume 
\begin{equation}\label{eq:U-delta}
  (U^+_r,U^-_r)=(\delta,1)
\end{equation}
with $\delta=0$ or 1.
The equation \eqref{eq:b2i} with $p=r+m$, $q-p=\pm r$ and $m<0$ means
$(r+m)(2r+m)U^{\pm}_rV_m = 0$
and therefore
\begin{equation}
\begin{aligned}
   V(t) &= at^{-2r} + bt^{-r} + \sum_{j>0}V_jt^j,\\
   r' &= 
    \begin{cases}
       -2r\quad&\text{if }a\ne 0,\\
       -r\quad&\text{if }a = 0.\\
    \end{cases}
\end{aligned}
\end{equation}
Here $(a,b)\ne(0,0)$.
Moreover \eqref{eq:b2i} with $p=r$ shows
\begin{equation}\label{eq:b2negW}
 U^+_rW_{q-2r} = U^-_rW_q\quad
\text{ if }q\notin\{-2r,-r,0,r,2r,3r,4r\}.
\end{equation}
Put $\bar r=-\max\{r',r''\}>0$.
Then \eqref{eq:b2i} with $q=-\bar r$ means
\begin{equation}\label{eq:b2U}
  p\bar r(2p+\bar r)(p+\bar r)(V_{-\bar r}U^-_{p+\bar r} - W_{-\bar r}U^-_p)=0.
\end{equation}
Similarly \eqref{eq:b2i} with $q=-r$ means
\begin{equation}\label{eq:b2-41}
  p\bar r(2p+r)(p+r)(bU^-_{p+r} - W_{-r}U^-_p)=0.
\end{equation}

If $r' > r''$, we have $U^-=0$ because $V_{-\bar r}=0$.
Hence $r'\le r''$ and we may moreover assume
\begin{equation}\label{eq:VW-delta}
  (V_{-\bar r}, W_{-\bar r})=(1,\epsilon) \quad\text{with }
    \epsilon=0\text{ or }1\quad\text{and}\quad
    \bar r=\begin{cases}
             2r&\text{if }a\ne0,\\
             r&\text{if }a=0.
           \end{cases}
\end{equation}
Then \eqref{eq:b2negW} implies
\[
 W(t) = e_1t^r(1-\delta t^r)^{-1} + e_2t^{2r}(1-\delta t^{2r})^{-1}
     + e_3t^{-2r}+ e_4t^{-r} + e_5t^{r} + e_6t^{2r} + e_7t^{3r} + e_8t^{4r}
\]
and it follows from \eqref{eq:U-delta}, \eqref{eq:b2U} and \eqref{eq:VW-delta} that
\begin{equation}\label{eq:U-minus}
 U^-(s) = s^r(1-\epsilon s^r)^{-1} + cs^{2r}(1-\epsilon s^{2r})^{-1}.
\end{equation}
If $\epsilon=1$, we may assume that $U^-(s)$ has a pole at $s=1$ as 
in the argument in Case 1-1 and then $b=W_r$ in \eqref{eq:b2-41}.  
Note that if $b\ne0$, \eqref{eq:b2-41} implies $c=0$.
Hence
\begin{equation}\label{eq:bc0}
 bc=0
\end{equation}
and \eqref{eq:b2-41} with $p=r$ means
\begin{equation}
 e_4=0\quad\text{if }\epsilon=0.
\end{equation}

{\bf Case 4-1}: $(e_1,e_2)\ne0$ and $\delta=1$.\newline
We may also assume $W(t)$ has a pole at $t=1$.
If $\epsilon=1$, then $U^-(t)$ and $W(t)$ have poles in $\mathbb C\setminus\{0\}$
and this case is reduced to Proposition~\ref{prop:rat}~iii).
Hence we may assume $\epsilon=0$ and therefore $e_3=0$ by \eqref{eq:VW-delta}.
Then Proposition~\ref{prop:rat}~ii) assures 
\begin{equation}
\begin{aligned}
 W(t)&=e_1t^r(1- t^r)^{-1} + e_2t^{2r}(1- t^{2r})^{-1},\\
  U^+(s)&=U^-(s)=s^r+cs^{2r}.
\end{aligned}
\end{equation}
Now \eqref{eq:b2i} with $(p,q)=(2r,r)$ means $ce_1=0$.
Thus $(U^+(t),U^-(t),V(t),0)\in\mathcal S(B_2)$ and
it follows from Case 1 that $V(t)=at^{-2r} + bt^{-r}$ with $bc=0$.
Then the solution corresponds to (Toda-$B_2^{(1)}$-bry) or (Toda-$B_2^{(1)}$-S-bry).

{\bf Case 4-2}. $e_1=e_2=0$, $\epsilon=1$.\newline
Proposition~\ref{prop:rat}~ii) assures 
$V(t)=W(t)=at^{-2r}+ bt^{-r} + e_5t^{r} + e_6t^{2r} + e_7t^{3r} + e_8t^{4r}$
with $bc=0$.
Putting $q=2p-{\bar r}$ in \eqref{eq:b2i} we have 
$V_{\bar r}U^+_{p-\bar r}+W_{-\bar r}U^+_p=0$ if $p$ is sufficiently large positive 
integer.  
Hence if $U^+(t)$ is not a polynomial of $t$, 
it has a pole in $\mathbb C\setminus\{0\}$ 
and this case is reduced to Proposition~\ref{prop:rat}~iii).

Thus we may assume $U^+(s)=\sum_{i=r}^NU^+_is^i$ with $U^+_N\ne0$.
Suppose $W_j=0$ for $j>M$.
If $M>0$, the coefficients of $s^Nt^{M+2N}$ in \eqref{eq:b2fun} implies 
$W_MU^+_N=0$.  Hence $e_5=e_6=e_7=e_8=0$ and therefore $(U^+,0,V,W)\in\mathcal S(B_2)$
and Case 1 implies $U^+(s)=s^r+c's^{2r}$ with $bc'=0$.
The solution is a standard transform of case V.

{\bf Case 4-3}: $e_1=e_2=\epsilon=0$.\newline
Then $U^-(s)=s^r + cs^{2r}$ and 
$W(t)=e_5t^{r} + e_6t^{2r} + e_7t^{3r} + e_8t^{4r}$.
Putting $p=q+r$ in \eqref{eq:b2i}, we have
$V_qU^-_r-W_qU^-_{q+r}=0$ for $q>0$ and therefore
$V(t)=at^{-2r}  + b t^{-r} + ce_5t^r$.

Suppose $U^+(s)$ is not a polynomial of $s$.
Putting $q=2p+\bar r$ in \eqref{eq:b2i}, we have 
$U^+_{p+\bar r}+W_{\bar r}U^+_p=0$
for a sufficiently large $p$.  
We may assume $U^+(s)$ has a pole at $s=1$.
Then $W_{\bar r}=-1$ and Proposition~\ref{prop:rat}~ii)
with Lemma~\ref{rem:b2} proves $V(t^{-1})+W(t)=0$
and $V(t)=at^{-2r}+bt^{-r}$ with $bc=0$.
Thus $(U^+,0,V,W)\in\mathcal S(B_2)$ and 
$U^+(s)=d_1s^r(1-s^{r})^{-1} + d_2s^{2r}(1-s^{2r})^{-1}$ 
with $bd_2=0$
and the solution is a standard transform of Case IV.

Now we may assume $U^+(s)=\sum_{i=r}^NU^+_is^i$ and
$W(t)=\sum_{i=r}^MW_it^i$ with $W_MU^+_N\ne0$.
Then \eqref{eq:b2i} 
with $(p,q)=(N, M+2N)$ shows $W_MU^+_N=0$, which contradicts
to the assumption.
\end{proof}
\begin{cor}\label{cor:regA}
The non-trivial solutions $R(x,y)$ of \eqref{eq:b2-0} with regular 
singularity at the point $t=0$ are transformations of the following solutions
under translations. % and symmetries.
\begin{align*}
&\begin{aligned}
 &C_1\bigl(\sh^{-2}\lambda(x+y) + \sh^{-2}\lambda(x-y)\bigr)
+C_2\bigl(\sh^{-2}\lambda x + \sh^{-2}\lambda y\bigr)\\
&\quad+C_3\bigl(\sh^{-2}2\lambda x + \sh^{-2}2\lambda y\bigr)
+C_0,
\end{aligned}\tag{\rm Trig-$BC_2$-reg}
\allowdisplaybreaks\\
&\begin{aligned}
 &C_1\bigl(\sh^{-2}\lambda(x+y) + \sh^{-2}\lambda(x-y)\bigr)\\
 &\quad+C_2\bigl(\sh^{-2}2\lambda(x+y) + \sh^{-2}2\lambda(x-y)\bigr)\\
 &\quad+C_3\bigl(\sh^{-2}2\lambda x + \sh^{-2}2\lambda y\bigr)
 +C_0,
\end{aligned}\tag{\rm Trig${}^d$-$BC_2$-reg}
\allowdisplaybreaks\\
&C_1\bigl(e^{-2\lambda(x+y)}+e^{-2\lambda(x-y)})
+C_2\sh^{-2}\lambda y +C_3\sh^{-2}2\lambda y
+C_0,\tag{\rm Toda-$D_2$-bry}
\\
&C_1\sh^{-2}\lambda(x-y) +C_2\sh^{-2}2\lambda(x-y)
+ C_3\bigl(e^{-4\lambda x}+e^{-4\lambda y})
+C_0,\tag{Toda${}^{d}$-$D_2$-bry}
\allowdisplaybreaks\\
&\begin{aligned}
 C_1\sh^{-2}\lambda(x-y)
+C_2\bigl(e^{-2\lambda x}+e^{-2\lambda y})
+C_3\bigl(e^{-4\lambda x}+e^{-4\lambda y})
+C_0,
\end{aligned}\tag{\rm Trig-$A_1$-bry-reg}
\allowdisplaybreaks\\
&\begin{aligned}
 &C_1\bigl(e^{-\lambda(x+y)}+e^{-\lambda(x-y)})
+C_2\bigl(e^{-2\lambda(x+y)}+e^{-2\lambda(x-y)})\\
&\quad+C_3\sh^{-2}\lambda y
+C_0,
\end{aligned}\tag{\rm Trig${}^d$-$A_1$-bry-reg}
\allowdisplaybreaks\\
 &C_1e^{-\lambda(x-y)} + C_2e^{-\lambda y} + C_3e^{-2\lambda y}+C_0,
\tag{\rm Toda-$BC_2$}\\
 &C_1e^{-\lambda(x-y)} + C_2e^{-2\lambda(x-y)} + C_3e^{-2\lambda y}+C_0
\tag{\rm Toda${}^d$-$BC_2$}.
\end{align*}
\end{cor}

\section{Type $B_n$ ($n\ge3$)}\label{sec:Bn}
Let $\mathbb R^n$ be the Euclidean space with the natural inner product
$\langle x,y \rangle=\sum_{i=1}^nx_iy_i$
for $x=(x_1,\ldots,x_n)$, $y=(y_1,\ldots,y_n)\in\mathbb R^n$.
Then $e_i=(\delta_{i1},\ldots,\delta_{i\nu},\ldots,\delta_{in})$ for $i=1,\ldots,n$
form a natural orthonormal basis of $\mathbb R^n$.
For $v\in\mathbb R^n$, let $\p_v$ be the differential operator defined by 
$(\p_v\psi)(x)=\dfrac{d\psi(x+tv)}{dt}\Big|_{t=0}$ for a function $\psi(x)$ on 
$\mathbb R^n$ and we put $\p_i=\p_{e_i}$.
If $v\ne0$, the reflection $w_v$ with respect to $v$ is a linear transformation
of $\mathbb R^n$ defined by $w_v(x)=x -\dfrac{2\langle v,x\rangle}{\langle v,v\rangle}v$
for $x\in\mathbb R^n$.

The root system $\Sigma=\Sigma(B_n)$ of type $B_n$ is realized in $\mathbb R^n$ by
\begin{equation}
\begin{cases}
 \Sigma(A_{n-1})^+ = \{e_i-e_j;\, 1 \le i<j\le n\},\\
 \Sigma(D_n)^+ = \Sigma_L^+ = \{e_i \pm e_j;\, 1\le i<j\le n\},\\
 \Sigma(D_n) = \Sigma_L = \{\alpha,-\alpha;\,\alpha\in\Sigma(D_n)^+\},\\
 \Sigma(B_n)_S^+ = \Sigma_S^+ = \{e_k;\, 1\le k\le n\},\\
 \Sigma(B_n)^+ = \Sigma^+ = \Sigma(D_n)^+\cup\Sigma(B_n)^+_S,\\
 \Sigma(B_n) = \{\alpha,\,-\alpha;\,\alpha\in\Sigma(B_n)^+\}.
\end{cases}
\end{equation}
The Weyl group $W_\Sigma$ of $\Sigma$ is the finite group generated by 
$\{w_\alpha;\,\alpha\in\Sigma\}$,
which is the group generated by the permutation of the coordinate 
$(x_1,\ldots,x_n)$ of 
$\mathbb R^n$ and by the change of the signs of some coordinates $x_i$.
For a subset $F$ of $\Sigma$, let $W_F$ denote the subgroup of $W_\Sigma$ generated by 
$\{w_\alpha;\,\alpha\in F\}$.  Then we call the set 
$\bar F=\{w\alpha;\,w\in W_F\text{ and }\alpha\in F\}$ the root system generated by 
$F$ and $W_F$ the Weyl group of the root system $\bar F$.
%%%%
Let
\begin{equation}
 P=\sum_{j=1}^n\frac{\p^2}{\p x_j^2} + R(x)
\end{equation}
be a differential operator with a function $R(x)$ such that it admits a differential operator
\begin{equation}
 Q=\sum_{j=1}^n \frac{\p ^4}{\p x_j^4} + S
 \qquad\text{with }\ord S<4
\end{equation}
satisfying $PQ=QP$.

Now we assume
\begin{equation}\label{eq:u-alpha}
\begin{aligned}
 R(x) &= \sum_{\alpha\in\Sigma(B_n)^+}u_\alpha(\langle\alpha,x\rangle)\\
      &= \sum_{1\le i<j\le n}\bigl(u^{ij}(x_i+x_j)+v^{ij}(x_i-x_j)\bigr)
         + \sum_{k=1}^n w^k(x_k),\\
 u^{ij}&= u_{e_i-e_j},\ v^{ij}=u_{e_i+e_j}\quad\text{and}\quad w^k=u_{e_k}\\
 &\quad\text{for }1\le i<j\le n\text{ and }1\le k\le n.
\end{aligned}
\end{equation}
For $\alpha\in\Sigma(B_n)^+$, 
we put $u_{-\alpha}(t)=u_\alpha(-t)$ for the convention.

Fix indices $i$ and $j$ with $1\le i<j\le n$ and put $u^{ji}(t) = u^{ij}(-t)$
and $I(i,j)=\{1,\ldots,n\}\setminus\{i,j\}$.
It follows from the proof of \cite[Theorem~6.1]{OOS} that
the condition for the existence of $Q$ is equivalent to
\begin{equation}\label{eq:bn}
S_{ij}=S_{ji}\qquad(1\le i<j\le n)
\end{equation}
with
\[
\begin{aligned}
S^{ij} &= \Bigl(\p_i^2 w^i(x_i) + \sum_{\nu\in I(i,j)}\p_i^2\bigl(u^{i\nu}(x_i+x_\nu)
 +v^{i\nu}(x_i-x_\nu)\bigr)\Bigr)\\
&\quad\cdot\Bigl(u^{ij}(x_i+x_j)-v^{ij}(x_i-x_j)\Bigr)\\
&+ 3\Bigl(\p_i w^i(x_i) + \sum_{\nu\in I(i,j)}
 \p_i\bigl(u^{i\nu}(x_i+x_\nu)+v^{i\nu}(x_i-x_\nu)\bigr)\Bigr)\\
&\quad\cdot\Bigl(\p_iu^{ij}(x_i+x_j)-\p_iv^{ij}(x_i-x_j)\Bigr)\\
&+ 2\Bigl(w^i(x_i) + 
\sum_{\nu\in I(i,j)}\bigl(u^{i\nu}(x_i+x_\nu)+v^{i\nu}(x_i-x_\nu)\bigr)\Bigr)\\
&\quad\cdot\Bigl(\p_i^2u^{ij}(x_i+x_j)-\p_i^2v^{ij}(x_i-x_j)\Bigr)\\
&+ \sum_{\nu\in I(i,j)}\Bigl(\p_i^2u^{i\nu}(x_i+x_\nu)-\p_i^2v^{i\nu}(x_i-x_\nu)\Bigr)
  \Bigl(u^{j\nu}(x_j+x_\nu) - v^{j\nu}(x_j-x_\nu)\Bigr).
\end{aligned}
\]

Then we have assumed that
\begin{equation}
  u_\alpha(\log t)=\sum u^\alpha_\nu t^\nu\quad\text{for }\alpha\in\Sigma^+
\end{equation}
with $u^\alpha_\nu\in\mathbb C$.
Here $u_\alpha(\log t)$ is analytic if $0<|t|\ll 1$ and $u^\alpha_\nu=0$
if $\nu$ is a sufficiently large negative integer.

Put
\begin{equation}
\begin{aligned}
 t_j &= e^{-x_j + x_{j+1}}\ (j=1,\ldots,n-1),\quad t_n = e^{-x_n},\\
 u^{ij}(x_i+x_j)&
   =\sum u^{ij}_\nu t_i^\nu\cdots t_{j-1}^\nu t_j^{2\nu}\cdots t_n^{2\nu},\\
 v^{ij}(x_i-x_j)&
   =\sum v^{ij}_\nu t_i^\nu\cdots t_{j-1}^\nu,\\
 w^i(x_k)&
   =\sum w^i_\nu t_k^\nu\cdots t_n^\nu,\\
  U_\alpha(t) &= \sum_{\nu\in\mathbb Z\setminus\{0\}} U^\alpha_\nu t^\nu\quad\text{with }
   u^\alpha_\nu = \nu U^\alpha_\nu\text{ and }\alpha\in\Sigma^+.
\end{aligned}
\end{equation}
Here $1\le i<j\le n$ and $u^{ij}_\nu$, $v^{ij}_\nu$ and $w^i_\nu\in\mathbb C$
and they are zero if $\nu$ is a sufficiently big negative integer.
Then the coefficients of $(t_i\cdots t_{j-1})^q(t_j\cdots t_n)^p$
in \eqref{eq:bn} show
\[
   pqw^i_{2p-q}u^{ij}_{q-p} - p(2p-q)w^j_qv^{ij}_{p-q}
 = q(q-p)w^i_{q-2p}u^{ij}_p - (2p-q)(p-q)w^ju^{ij}_p
\]
if $pq(p-q)(2p-q)\ne0$ and therefore by putting
\begin{equation}
  \begin{cases}
    U^\pm(t) = \sum_{\nu\ge r}U^\pm_it^\nu,\ 
    V(t) = \sum_{\nu\ge r}V_\nu t^\nu \text{ and }
    W(t) = \sum_{\nu\ge r''}W_\nu t^\nu,\\
    U^\pm_0=V_0=W_0=0,\\
    u^{ij}(t)=t(U^+)'(t)+u^{ij}_0,\ v^{ij}(t)=t(V^-)'(t)+v^{ij}_0,\\
    w^i(t)=tV'(t)+w^i_0
    \text{ and }w^j(t)=tW'(t)+w^j_0,
  \end{cases}
\end{equation}
we have \eqref{eq:b2i} and \eqref{eq:b2fun}.
Hence $(u^{ij}, v^{ij}, w^i, w^j)$ is a standard transformation of a solution
of type $B_2$ studied in \S\ref{sec:B2}.

Suppose $\{\alpha,\beta,\alpha+\beta\}\subset\Sigma^+_L$.
Then $(\alpha,\beta,\alpha+\beta)$ is one of the followings
\begin{gather}
  (e_{i_1-i_2}, e_{i_2-i_3}, e_{i_1-i_3}),\label{eq:d3-1}\\
  (e_{i_1-i_2}, e_{i_2+i_3}, e_{i_1+i_3}),\label{eq:d3-2}\\
  (e_{i_2-i_3}, e_{i_1+i_3}, e_{i_1+i_2}),\label{eq:d3-3}\\
  (e_{i_1-i_3}, e_{i_2+i_3}, e_{i_1+i_2}) \label{eq:d3-4}
\end{gather}
with $1\le i_1<i_2<i_3\le n$.

Put $s=e^{-\langle\alpha,x\rangle}$ and $t=e^{-\langle\beta,x\rangle}$.
Moreover put $(i,j,\nu)=(i_1,i_2,i_3)$, $(i_1,i_2,i_3)$, $(i_2,i_3,i_1)$ and $(i_1,i_3,i_2)$
according to \eqref{eq:d3-1}, \eqref{eq:d3-2}, \eqref{eq:d3-3} and \eqref{eq:d3-4},
respectively.
Then $(u_\alpha,u_\beta,u_{\alpha+\beta})= (v^{ij},v^{j\nu},v^{i\nu})$, 
$(v^{ij},u^{j\nu},u^{i\nu})$, $(v^{ij},u^{j\nu},u^{i\nu})$ and 
$(v^{ij},u^{j\nu},u^{i\nu})$, respectively, and
the coefficients of $s^pt^q$ in \eqref{eq:bn} show
\begin{multline}
  (- q^2-3(p-q)q-2(p-q)^2)u^{\alpha+\beta}_qu^\alpha_{p-q} 
  + p^2u^{\alpha+\beta}_pu^\beta_{q-p}\\
 = (-q^2+3pq-2p^2)u^\beta_qu^\alpha_p + (q-p)^2u^{\alpha+\beta}_pu^\beta_{q-p}.
\end{multline}
if $pq(p-q)\ne0$.
Hence \begin{equation}\label{eq:bni1}
pq(p-q)(2p-q)(U^{\alpha}_pU^{\beta}_q - U^{\alpha}_{p-q}U^{\alpha+\beta}_q 
  - U^{\beta}_{q-p}U^{\alpha+\beta}_p)=0.
\end{equation}
Now put $(i,j,\nu)=(i_2,i_3,i_1)$, $(i_1,i_3,i_1)$, $(i_1,i_3,i_2)$ and $(i_2,i_3,i_1)$
according to \eqref{eq:d3-1}, \eqref{eq:d3-2}, \eqref{eq:d3-3} and \eqref{eq:d3-4}, 
respectively.
Then $(u_\alpha,u_\beta,u_{\alpha+\beta})= (v^{i \nu},v^{ij},v^{j\nu})$, 
$(v^{i\nu},u^{ij},u^{j\nu})$, $(v^{j\nu},u^{ij},u^{i\nu})$ and 
$(v^{j\nu},u^{ij},u^{i\nu})$, respectively,
and the coefficients of $s^pt^q$ in \eqref{eq:bn} show
\begin{multline}
  \bigl(p^2+3(q-p)p+2(q-p)^2\bigr)u^{\alpha+\beta}_pu^\beta_{q-p} 
  - q^2u^{\alpha+\beta}_qu^\alpha_{p-q}\\
 = (p^2-3pq+2q^2)u^\alpha_pu^\beta_q - (p-q)^2u^{\alpha+\beta}_qu^\alpha_{p-q}
\end{multline}
if $pq(p-q)\ne0$ and we have
\begin{equation}\label{eq:bni2}
pq(p-q)(p-2q)(U^{\alpha}_pU^{\beta}_q - U^{\alpha}_{p-q}U^{\alpha+\beta}_q 
  - U^{\beta}_{q-p}U^{\alpha+\beta}_p)=0.
\end{equation}
Combining \eqref{eq:bni1} and \eqref{eq:bni2}, we get
\begin{equation}\label{eq:bni0}
pq(p-q)(U^{\alpha}_pU^{\beta}_q - U^{\alpha}_{p-q}U^{\alpha+\beta}_q 
  - U^{\beta}_{q-p}U^{\alpha+\beta}_p)=0.
\end{equation}
Namely,
\begin{equation}\label{eq:A2B}
 \bigl(U_\alpha(s)+U_\beta(t)-U_{\alpha+\beta}(st)\bigr)^2=
  F_\alpha(s)+F_\beta(t)+F_{\alpha+\beta}(st)
\end{equation}
with suitable functions $F_\alpha$, $F_\beta$ and $F_{\alpha+\beta}$.
Then if at least two in $\{U_\alpha,U_\beta,U_{\alpha+\beta}\}$ do not
vanish, Proposition~\ref{prop:A} shows that
$\bigl(U_\alpha(t),U_\beta(t),U_{\alpha+\beta}(t)\bigr)$ is a standard 
transformation of 
$\bigl(t^r(1-t^r)^{-1},t^r(1-t^r)^{-1},t^r(1-t^r)^{-1}\bigr)$ or 
$(C_1t^r,C_2t^r,C_3t^{-r})$.

The argument above shows the following lemma.

\begin{lem}\label{lem:two}
Let $\alpha$ and $\beta\in\Sigma$ such that
$\alpha\ne\pm\beta$, $\langle\alpha,\beta\rangle\ne0$, $|\alpha|\ge|\beta|$,
$U_\alpha\ne0$ and $U_\beta\ne0$.
Suppose
\[
  U_\alpha(t)=C_1t^r.
\]
Then we have the following two cases.

{\bf Case 1:} $|\alpha|=|\beta|$.%\newline
\begin{equation*}
 U_\beta(t) =
 \begin{cases}
   C_1't^r\quad&\text{if }\langle\alpha,\beta\rangle<0,\\
   C_1't^{-r}\quad&\text{if }\langle\alpha,\beta\rangle>0.
 \end{cases}
\end{equation*}

{\bf Case 2:} $|\alpha|^2=2|\beta|^2$.%\newline
\begin{equation*}
   U_\beta(t)=C'_1t^r(1-t^r)+C'_2t^{2r}(1-t^{2r})\text{ and }
   U_{w_\beta(\alpha)}(t)=U_\alpha(t)\text{ under a translation}
\end{equation*}
or
\begin{equation*}
 U_\beta(t)=
 \begin{cases}
   C_1't^r+C_2't^{2r}\quad&\text{if }\langle\alpha,\beta\rangle<0,\\
   C_1't^{-r}+C_2't^{-2r}\quad&\text{if }\langle\alpha,\beta\rangle>0.
 \end{cases}
\end{equation*}
\end{lem}

\begin{defn}\label{def:comp}
For the functions $u_\alpha$ in \eqref{eq:u-alpha}, put
\begin{equation}
 \Delta = \{\alpha\in\Sigma(B_n)^+;\, u'_\alpha\ne0\}.
\end{equation}
Let $\bar\Delta$ be the root system generated by $\Delta$
and let
\begin{equation}
 \bar\Delta = \bar\Delta_1\cup\cdots\cup\bar\Delta_N
\end{equation}
be the decomposition into irreducible root systems.
Put
\begin{equation}
  \Delta_k = \bar\Delta_k\cap\Delta
\end{equation}
and we call it an {\it irreducible component\/} of $\Delta$.

We say that $P$ with the potential function $R(x)$ is {\it irreducible\/} if 
$\bar\Delta$ is an irreducible root system of rank $n$ or of type $A_{n-1}$.
\end{defn}
\begin{rem} {\rm i)}
$\Delta=\Delta_1\cup\cdots\cup\Delta_N$.

{\rm ii)}
 Suppose $\alpha\in\Delta_k$.  Then $\beta\in\Delta$ is an element of $\Delta_k$
if and only if there exists a sequence
$\alpha_1=\alpha$, $\alpha_2,\ldots,\alpha_\ell=\beta\in\Delta$ such that
$\langle\alpha_i,\alpha_{i+1}\rangle\ne0$ for $i=1,\ldots,\ell-1$.
\end{rem}

\begin{lem}\label{lem:long}
Let $\Delta'$ be an irreducible component of $\Delta$
and let $\bar\Delta'$ and $\bar\Delta'_L$ be the root systems generated 
by $\Delta'$ and $\Delta'_L:=\Delta'\cap\Sigma_L$, respectively.
Suppose $\bar\Delta'$ is of type $B_m$ with $m>2$.
Then $\bar\Delta'_L$ is of type $A_{m-1}$ or type $D_m$.
\end{lem}
\begin{proof}
Since $\Delta'_L$ and $\{e_1,\ldots,e_m\}$ generate $\Delta'$,
there exist $\alpha_i\in\Delta'_L$ such that
$\langle\alpha_i,e_i\rangle\ne0$ and 
$\langle\alpha_i,e_{i+1}\rangle\ne0$ for $1\le i< m$.
Since $\alpha_1,\ldots,\alpha_{m-1}$ generate a root system of type $A_{m-1}$,
$\bar\Delta'_L$ is of type $A_{m-1}$ or type $D_m$.
\end{proof}
\begin{lem}\label{lem:EXT}
Let $S$ be a subset of a classical root system $\tilde\Sigma$ of type $A$ or $B$.
Suppose $S$ generates an irreducible root system $\bar S$ and
\[
 \langle\alpha,\beta\rangle\le0\quad\text{for }\alpha\in S\text{ and }\beta\in S
 \text{ with }\alpha\ne\beta.
\]
If the rank of $\bar S$ is larger than one, 
$S$ is the image of one of the following sets under the suitable 
transformation by an element of the Weyl group of $\tilde\Sigma$.
\begin{align}
 &\{e_1-e_2,\ldots,\,e_{m-1}-e_m\}\quad&(m\ge 3),\label{eq:Am}\\
 &\{e_1-e_2,\ldots,\,e_{m-1}-e_m,e_m\}\quad&(m\ge 2),\label{eq:Bm}\\
 &\{e_1-e_2,\ldots,\,e_{m-1}-e_m,\,e_{m-1}+e_m\}\quad&(m\ge4),\label{eq:Dm}\\
 &\{e_1-e_2,\ldots,\,e_{m-1}-e_m,\,e_m-e_1\}\quad&(m\ge 3),\label{eq:EAm}\\
 &\{e_1-e_2,\ldots,\,e_{m-1}-e_m,\,e_m,\,-e_1\}\quad&(m\ge 2),\label{eq:EBm}\\
 &\{e_1-e_2,\ldots,\,e_{m-1}-e_m,\,e_{m-1}+e_m,
\,-e_1-e_2\}\quad&(m\ge 4),\label{eq:EDm}\\
 &\{e_1-e_2,\ldots,\,e_{m-1}-e_m,\,e_{m-1}+e_m,\,-e_1\}\quad&(m\ge 3).\label{eq:BDm}
\end{align}
\end{lem}
\begin{proof}
Let $S'$ be a subset of $S$ such that
$S'$ is a transformation of one of the sets $\{e_1\}$, $\{e_1-e_2\}$, 
\eqref{eq:Am}, \eqref{eq:Bm} and \eqref{eq:Dm} by an element of
the Weyl group of $\tilde\Sigma$.

If the number of the elements of $S'$ is smaller than the rank of $\bar S$,
there exists $\alpha\in S$ and $\beta\in S'$ such that
$\langle\alpha,\beta\rangle\ne0$ and 
$\alpha\notin\sum_{\gamma\in S'}\mathbb R\gamma$.
Then it is easy to see that $S'\cup\{\alpha\}$ is a transformation
of \eqref{eq:Am}, \eqref{eq:Bm} or \eqref{eq:Dm} by a suitable
element of the Weyl group.

Thus we may assume that $S'$ equals \eqref{eq:Am} or \eqref{eq:Bm} or \eqref{eq:Dm}
and that the number of the elements of $S'$ equals the rank of $\bar S$.
Put $S_o=\{\beta\in\tilde\Sigma\cap\sum_{\gamma\in S'}\mathbb R\gamma;
\langle\beta,\gamma\rangle\le0\ \text{for any }\gamma\in S'\}$.
Then if $S'$ is \eqref{eq:Am} or \eqref{eq:Bm}, 
$S_o=\{e_m-e_1\}$ or $\{-e_1, -e_1-e_2\}$, respectively.
If $S'$ is \eqref{eq:Dm}, then 
$S_o=\{-e_1-e_2,-e_1\}$ or $\{-e_1-e_2\}$ according to
$\tilde\Sigma$ is of type $B$ or $A$, respectively.
Thus the lemma is clear because $S'\subset S\subset S'\cup S_o$.
\end{proof}

\begin{lem}
Let $\bar\Delta$ be a root system generated by a classical root system $\tilde\Sigma$
of type $A_{n-1}$ or $B_n$.
Let $\bar\Delta=\bar\Delta_1\cup\cdots\cup\bar\Delta_{N'}\cup\cdots\cup\bar\Delta_N$
be an irreducible decomposition so that the rank of $\bar\Delta_i$ is larger 
than one for $1\le i\le N'$ and the rank of $\bar\Delta_i$ equals one for $N'<i\le N$.
Then under the transformation by an element of the Weyl group of $\Sigma$,
there exists a sequence of integers $0=n_0<n_1<n_2<\cdots<n_{N'}$ such that
$\bar\Delta_i$ is generated by
\begin{equation*}
 \begin{cases}
    \{e_{n_{i-1}+1}-e_{n_{i-1}+2},\ldots,\,e_{{n_i}-1}-e_{n_i}\}
       \quad&\text{if $\bar\Delta_i$ is of type A},\\
    \{e_{n_{i-1}+1}-e_{n_{i-1}+2},\ldots,\,e_{{n_i}-1}-e_{n_i},\, e_{{n-i}-1}+e_{n_i}\}
        \quad&\text{if $\bar\Delta_i$ is of type D},\\
        \{e_{n_{i-1}+1}-e_{n_{i-1}+2},\ldots,\,e_{{n_i}-1}-e_{n_i},\, e_{n_i}\}
        \quad&\text{if $\bar\Delta_i$ is of type B}
 \end{cases}
\end{equation*}
for $1\le i\le N'$. Moreover if $N'<i\le N$.
$\bar\Delta_i$ equals $\{\pm e_\nu\}$, $\{\pm (e_\nu-e_{\nu+1})\}$ or
$\{\pm (e_\nu+e_{\nu+1})\}$ for a suitable $\nu$ with $\nu>n_{N'}$.
\end{lem}
\begin{proof}
Note that $\{\alpha\in\tilde\Sigma;\,\langle e_1-e_2,\alpha\rangle=0\}$
is generated by 
\[
 \begin{cases}
 \{e_3-e_4,\ldots,\,e_{n-1}-e_n\}\quad&\text{if $\tilde\Sigma$ is of type $A_n$},\\
 \{e_3-e_4,\ldots,\,e_{n-1}-e_n,\,e_n\}\text{ and }\{e_1+e_2\}&\text{if $\tilde\Sigma$ is of type $B_n$}.
 \end{cases}
\]
Hence  the lemma is clear by the induction on $N$ if $N'=0$.

Suppose $N'>0$.
By the preceding lemma, we may assume that the fundamental system of $\Delta_1$ is
\eqref{eq:Am}, \eqref{eq:Bm} or \eqref{eq:Dm}.
Then $\{\alpha\in\tilde\Sigma;\,\langle e_1-e_2,\alpha\rangle=0\}$ is generated by
\[
  \begin{cases}
 \{e_{m+1}-e_{m+2},\ldots,\,e_{n-1}-e_n\}\quad&\text{if $\tilde\Sigma$ is of type $A_n$},\\
 \{e_{m+1}-e_{m+2},\ldots,\,e_{n-1}-e_n,\,e_n\}&\text{if $\tilde\Sigma$ is of type $B_n$}
 \end{cases}
\]
and the lemma is clear by the induction on $N'$.
\end{proof}

\begin{rem}\label{rm:bn}
i) Fix $\alpha\in\Delta'$.
Let $v\in\mathbb R^n$ with $\langle\alpha,v\rangle=0$.
Then $\p_vu_\alpha(\langle\alpha,x\rangle)=0$.

ii) If the rank of $\bar\Delta'$ equals one, 
$u_\alpha(t)$ for $\alpha\in\Delta'$ is any function.

iii) If the rank of $\bar\Delta'$ is larger than one, $U_\alpha(t)$ with $\alpha\in\Delta'$ 
are global meromorphic functions 
and therefore we may study $\{U_\alpha(t);\,\alpha\in\Delta'\}$
under the image of a transformation by the Weyl group.

iv) By the irreducible decomposition in Definition~\ref{def:comp} 
our classification reduces to the case when $P$ is irreducible.
\end{rem}
\begin{thm}\label{prop:bn}
Let $\Delta'$ be an irreducible  component of $\Delta$.
Then the potential function
$R_{\Delta'}(x):=\sum_{\alpha\in\Delta'}u_{\alpha}(\langle\alpha,x\rangle)$ 
is a transformation of a function in the following list with $m\ge2$ under 
a map generated by the Weyl group, translations and scalings of the 
coordinates (cf.~Lemma~\ref{rem:b2}).
\begin{gather*}
\intertext{{\rm Type $A_1$:}
If the rank of $\bar\Delta'$ equals $1$, 
$R_{\Delta'}(x)$ is an arbitrary function of $\langle\alpha,x\rangle$
with $\alpha\in\Delta'$.
This solution is called trivial.}
\intertext{{\rm Type $B_2$:}
A standard transform of the function in the list {\rm (Trig-$B_2$) -- (Toda-$C_2^{(1)}$)} 
in $\S\ref{sec:B2}$ with replacing $(x,y)$ by $(x_1,x_2)$.}
 \begin{split}
  \intertext{{\rm(Trig-$B_m$):} Trigonometric potential of type $B_m$:}
  &\sum_{1\le i<j\le m}C_0\bigl(\sh^{-2}\lambda(x_i+x_j)
  +\sh^{-2}\lambda(x_i-x_j)\bigr)\\
  &\quad+\sum_{k=1}^m\bigl(
    C_1\sh^{-2}2\lambda x_k + C_2\sh^{-2}\lambda x_k + C_3\ch2\lambda x_k
  + C_4\ch4\lambda x_k\bigr).
  \end{split}\label{eq:BnP}\allowdisplaybreaks\\
%%%
 \begin{split}
\intertext{{\rm(Trig-$A_{m-1}$-bry):} Trigonometric potential of type $A_{m-1}$ with boundary:}
 &\sum_{1\le i<j\le m}C_0\sh^{-2}\lambda(x_i-x_j)\\
 &\quad+ \sum_{k=1}^m\bigl(C_1e^{-2\lambda x_k} + 
   C_2e^{-4\lambda x_k}+C_3e^{2\lambda x_k} + C_4e^{4\lambda x_k}\bigr),
 \end{split}\label{eq:Anb}\allowdisplaybreaks\\
%%%
 \begin{split}
\intertext{{\rm(Toda-$B_m^{(1)}$-bry):} Toda potential of type $B_m^{(1)}$ with boundary:}
 &\sum_{i=1}^{m-1}C_0e^{-2\lambda(x_i-x_{i+1})}
 + C_0e^{-2\lambda(x_{m-1}+x_m)}
 + C_1e^{2\lambda x_1} + C_2e^{4\lambda x_1}\label{eq:EBnb}\\
 &\quad+ C_3\sh^{-2}\lambda x_m +  C_4\sh^{-2}2\lambda x_m,
 \end{split}\allowdisplaybreaks\\
%%%
\intertext{{\rm(Toda-$C_m^{(1)}$):} Toda potential of type $C_m^{(1)}$:}
 \sum_{i=1}^{m-1}C_0e^{-2\lambda(x_i-x_{i+1})}
 + C_1e^{2\lambda x_1} + C_2e^{4\lambda x_1}
 + C_3e^{-2\lambda x_m}+ C_4e^{-4\lambda x_m},\label{eq:ECn}
\allowdisplaybreaks\\
%%%
 \begin{split}
\intertext{{\rm(Toda-$D_m^{(1)}$-bry):} Toda potential of type $D_m^{(1)}$ with boundary:}
 &\sum_{i=1}^{m-1}C_0\bigl(e^{-2\lambda(x_i-x_{i+1})}
 + e^{-2\lambda(x_{m-1}+x_m)}
 + e^{2\lambda(x_1+x_2)}\bigr)\label{eq:EDnb}\\
 &\quad+ C_1\sh^{-2}\lambda x_m + C_2\sh^{-2}{2\lambda x_m}
  + C_3\sh^{-2}\lambda x_1 + C_4\sh^{-2}{2\lambda x_1},
 \end{split}
%%%
\intertext{{\rm(Toda-$A_{m-1}^{(1)}$):} Toda potential of type $A_{m-1}^{(1)}$:}
 \sum_{i=1}^{m-1}C_0e^{-2\lambda(x_i-x_{i+1})}
 + C_0e^{2\lambda (x_1-x_m)}\label{eq:EAn}.
\end{gather*}
\end{thm}
%%%%
\begin{defn} We define some potential functions as specializations of the above.
 
(Trig-$A_{m-1}$): Trigonometric potential of type $A_{m-1}$ is (Trig-$A_{m-1}$-bry) with 
$C_1=C_2=C_3=C_4=0$.

(Toda-$B_m^{(1)}$): Toda potential of type $B_m^{(1)}$ is (Toda-$B_m^{(1)}$-bry) with $C_3=C_4=0$.

(Toda-$D_m^{(1)}$): Toda potential of type $D_m^{(1)}$ is (Toda-$D_m^{(1)}$-bry) with 
$C_1=C_2=C_3=C_4=0$.

(Toda-$D_m$-bry): Toda potential of type $D_m$ with boundary is (Toda-$B_m^{(1)}$-bry) with 
$C_1=C_2=0$.

(Toda-$A_{m-1}$): Toda potential of type $A_{m-1}$ is (Toda-$C_m^{(1)}$) with 
$C_1=C_2=C_3=C_4=0$.

(Toda-$BC_m$): Toda potential of type $B_m$ is (Toda-$C_m^{(1)}$) with $C_1=C_2=0$.

(Toda-$D_m$): Toda potential of type $D_m$ is (Toda-$B_m^{(1)}$-bry) with 
$C_1=C_2=C_3=C_4=0$.
\end{defn}
%%%%
{\em Proof of Proposition~\ref{prop:bn}}.
We may assume that $\bar\Delta'$ is not of type $B_2$. 
Then Lemma~\ref{lem:long} says that for any elements $\alpha$ and $\beta$ of 
$\Delta'_L$, there exists a sequence $\alpha=\alpha_1,\alpha_2,\ldots,\alpha_k=\beta$ 
such that $\langle\alpha_i,\alpha_{i+1}\rangle\ne0$ and $\alpha_i\in\Delta'_L$ for 
$i=1,\ldots,m-1$.
Note that the number of elements of $\Delta'_L$ is larger than one.
Fix $\alpha\in\Delta'$.
Then lemma~\ref{lem:two} assures that 
$U_\alpha(t)=Ct^r(1-at^r)^{-1}$ with $a\in\mathbb C$.

{\bf Case 1}: $U_{\alpha}(t)=Ct^r$.\newline
Lemma~\ref{lem:two} proves that $U_{\beta}(t)=C_\beta t^{2\epsilon(\beta)r}$ 
for $\beta\in\Delta'_L$.  Here $\epsilon(\beta)=1$ or $-1$.
Then the set $S_L=\{\epsilon(\beta)\beta;\,\beta\in\Delta'_L\}$ satisfies the 
assumption of Lemma~\ref{lem:EXT} and therefore we may assume that $S_L$ equals 
\eqref{eq:Am}, \eqref{eq:Dm}, \eqref{eq:EAm} or \eqref{eq:EDm}
under the transformation of an element of the Weyl group, which correspond to 
(Toda-$A_{m-1}$), (Toda-$D_m$), (Toda-$A_{m-1}^{(1)}$) and 
(Toda-$D_m^{(1)}$), respectively, if $U_{e_i}(t)=0$ for $i=1,\dots,m$.
Suppose $U_{e_i}(t)\ne0$ with a suitable $i$ satisfying $1\le i\le m$.
Then Lemma~\ref{lem:two} shows that one of the the following two cases occurs,
from which the statement of the proposition is clear.

{\bf Case 1-1}: $U_{e_i}(t)=C't^r(1-at^r)+C''t^{2r}(1-at^{2r})$ with $a\ne0$.\newline
We may assume $a=1$ by a translation.
Lemma~\ref{lem:two} shows that 
then $U_{-e_i-e_{i+1}}(t)=U_{e_i-e_{i+1}}(t)$ and if $i>1$
and $U_{e_{i-1}-e_i}(t)=U_{e_{i-1}+e_i}(t)$ if $i<m$.
Therefore %$i=1$ or $i=m$ and.
\[
 \begin{cases}
 i=1\text{ and $S_L$ equals \eqref{eq:EDm}}
      &\bigl(\Rightarrow\text{(Toda-$D_m^{(1)}$-bry)}\bigr)\\
  \text{\ or}\\
 i=m\text{ and $S_L$ equals \eqref{eq:Dm} or \eqref{eq:EDm}}
      &{\bigl(\Rightarrow\text{(Toda-$B_m^{(1)}$-bry) or (Toda-$D_m^{(1)}$-bry)}\bigr)}.\\
 \end{cases}
\]
\quad
{\bf Case 1-2}: $U_{e_i}(t)=C't^{\epsilon_ir}+C''t^{2\epsilon_ir}$
with $\epsilon_i=\pm1$.

\noindent
Lemma~\ref{lem:two} says $\epsilon_i\langle e_i,\alpha\rangle\le 0$ 
for $\alpha\in S_L$, only the following cases occur.
\[
 \begin{cases}
   i=1\text{ and $S_L$ equals \eqref{eq:Am} or \eqref{eq:Dm}}
   &{\bigl(\Rightarrow\text{(Toda-$C_m^{(1)}$) or (Toda-$B_m^{(1)}$-bry)}\bigr)},\\
   i=m\text{ and $S_L$ equals \eqref{eq:Am}}
   &{\bigl(\Rightarrow\text{(Toda-$C_m^{(1)}$)}\bigr)}.
 \end{cases}
\]
\quad
{\bf Case 2}: $U_\alpha(t)=C_\alpha t^r(1-a_\alpha t^r)^{-1}$ with $a_\alpha\ne0$.\newline
The argument just before Lemma~\ref{lem:two} says 
that the condition $U_\beta(t)\ne0$ and 
$U_\gamma(t)\ne0$ with $|\alpha|=|\beta|=|\gamma|$ means $U_{w_\beta(\gamma)}(t)\ne0$.
Hence $\{\pm\beta;\,\beta\in\Delta'_L\}$ is a root system of type $A_{m-1}$ or $D_m$.
We may assume $a_\alpha=1$ for its simple root $\alpha$
and hence $C_\alpha$ and $a_\alpha$ does not depend on $\alpha\in\Delta'_L$
because of \eqref{eq:A2B} and Proposition~\ref{rem:a2}.

{\bf Case 2-1}: $\bar\Delta'_L$ is of type $A$.\newline
Considering $\{U_{e_i+e_{i+1}}, U_{e_i-e_{i+1}}, U_{e_i}, U_{e_{i+1}}\}$, 
Theorem~\ref{thm:B2} shows 
$U_{e_i}(t)=U_{e_{i+1}}(t)=C_1t^{r}+C_2t^{2r}+C_3t^{-r}+C_4t^{-2r}$
and this case is reduced to (Trig-$A_{m-1}$-bry).

{\bf Case 2-2}: $\bar\Delta'_L$ is of type $D$.\newline
By the same consideration as in the previous case we may assume
$U_{e_\nu}(t)=C_1t^r(1-t^r)^{-1}+C_2t^{2r}(1-t^{2r})^{-1} + C_3(t^r-t^{-r})+C_4(t^{2r}-t^{-2r})$ for $\nu=1,\ldots,m$.
Hence this case is reduced to (Trig-$B_m$).
\qed
%\end{proof}

%%%%%%%%%%%%%%%%%%%%
\begin{cor}\label{cor:regB}
The non-trivial solutions in Proposition~\ref{prop:bn} which have regular 
singularity at the point $t=0$ are in Corollary~\ref{cor:regA} or 
in the following list.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{gather*}
\allowdisplaybreaks
 \begin{aligned}
  &\sum_{1\le i<j\le m}C_0\bigl(\sh^{-2}\lambda(x_i+x_j)
  +\sh^{-2}\lambda(x_i-x_j)\bigr)\\
  &\quad+\sum_{k=1}^m\bigl(
    C_1\sh^{-2}2\lambda x_k + C_2\sh^{-2}\lambda x_k\bigr),
  \end{aligned}
\tag{\rm Trig-$BC_m$-reg}
%%%
\allowdisplaybreaks\\
 \sum_{1\le i<j\le m}C_0\sh^{-2}\lambda(x_i-x_j)
 + \sum_{k=1}^m\bigl(C_1e^{-2\lambda x_k} + 
   C_2e^{-4\lambda x_k}\bigr),
\tag{\rm Trig-$A_{m-1}$-bry-reg}
%%%
\allowdisplaybreaks\\
 C_0\sum_{i=1}^{m-1}e^{-2\lambda(x_i-x_{i+1})}
 + C_0e^{-2\lambda(x_{m-1}+x_m)}
 + C_3\sh^{-2}\lambda x_m +  C_4\sh^{-2}2\lambda x_m,
\tag{\rm Toda-$D_m$-bry}\\
%%%
 C_0\sum_{i=1}^{m-1}e^{-2\lambda(x_i-x_{i+1})}
 + C_3e^{-2\lambda x_m}+ C_4e^{-4\lambda x_m}.
\tag{\rm Toda-$BC_m$}
\end{gather*}
\end{cor}
\begin{rem}\label{rm:reg}
We have a natural compactification $X$ of the space $\mathbb C^n$ of $t$
so that for every $w\in W_{\Sigma(B_n)}$ 
\[
 s^w_j = e^{-(x'_j-x'_{j+1})}\quad(j=1,\ldots,n-1),\quad s^w_n = e^{-x'_n}
 \quad\text{with }
 x'=wx
\]
gives a local coordinate system of $X$ and $t_j=s^e_j$ $(j=1,\dots,n)$.
Then the non-trivial potential functions $R(x)$ we have obtained is 
meromorphic on $X$.

If $R(x)$ is holomorphic at $(s^w_1,\dots,s^w_n)=0$ for any $w\in W_{\Sigma(B_n)}$, $R(x)$ is said to have {\it regular singularity at every infinity\/}.
In this case, our classification says that $R(x)$ is decomposed to the 
functions {\rm (Trig-$BC_m$-reg)} and {\rm (Trig-$A_m$)} which exactly
corresponds to Heckman-Opdam's potential function of classical type.
This gives a characterization of Heckman-Opdam's hypergeometric equations.
\end{rem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{document}

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