Some Remarks on the Shifner-Erougin-Salakhova-Chebotarev Type Differential Equations

Abstract

This article studies a type of integrable linear ordinary differential equations considered by Shifner [17] and generalised by Erougin [5], Salakhova and Chebotarev [15]. We will consider a slight generalisation of their results.

Appendix

In the appendix, we show Lemma 11 used in the main part.

Lemma 13

Let $A_{k}=\left[ x_{j-i+1}^{\left( k\right) }+\left( i-1\right) \alpha _{j-i+1}^{\left( k\right) }\right] _{i\le j}$ for $k=1,2$, where $\alpha _{\ell }^{\left( k\right) },x_{\ell }^{\left( k\right) }$ are arbitrary numbers, and $\left[ z_{ij}\right] _{i,j}=\left[ A_{1},A_{2}\right]$. Then,

$$\begin{aligned} z_{ij}&=\sum _{\ell =1}^{j-i+1}\left[ \left( \ell -1\right) \alpha _{j-i-\ell +2}^{\left( 2\right) }x_{\ell }^{\left( 1\right) }-\left( j-i-\ell +1\right) \alpha _{\ell }^{\left( 1\right) }x_{j-i-\ell +2}^{\left( 2\right) }\right] \nonumber \\&\quad +\left( i-1\right) \sum _{\ell =1}^{j-i+1}\left( i-j+2\ell -2\right) \alpha _{\ell }^{\left( 1\right) }\alpha _{j-i-\ell +2}^{\left( 2\right) }, \end{aligned}$$

(20)

for all i, j such that $i\le j$ (obviously $z_{ij}=0$ for $i>j$). In particular,

$$\begin{aligned} z_{1j}=\sum _{\ell =1}^{j}\left[ \left( \ell -1\right) \alpha _{j-\ell +1}^{\left( 2\right) }x_{\ell }^{\left( 1\right) }-\left( j-\ell \right) \alpha _{\ell }^{\left( 1\right) }x_{j-\ell +1}^{\left( 2\right) }\right] . \end{aligned}$$

(21)

Proof

For $i\le j$, one can easily compute

$$\begin{aligned} z_{ij}&=\left[ \sum _{\ell =i}^{j}\left\{ x_{\ell -i+1}^{\left( 1\right) }+\left( i-1\right) \alpha _{\ell -i+1}^{\left( 1\right) }\right\} \left\{ x_{j-\ell +1}^{\left( 2\right) }+\left( \ell -1\right) \alpha _{j-\ell +1}^{\left( 2\right) }\right\} \right] \\&-\left[ \sum _{\ell =i}^{j}\left\{ x_{\ell -i+1}^{\left( 2\right) }+\left( i-1\right) \alpha _{\ell -i+1}^{\left( 2\right) }\right\} \left\{ x_{j-\ell +1}^{\left( 1\right) }+\left( \ell -1\right) \alpha _{j-\ell +1}^{\left( 1\right) }\right\} \right] \\&=\sum _{\ell =i}^{j}\left[ \left( \ell -1\right) \alpha _{j-\ell +1}^{\left( 2\right) }x_{\ell -i+1}^{\left( 1\right) }-\left( i-1\right) \alpha _{\ell -i+1}^{\left( 2\right) }x_{j-\ell +1}^{\left( 1\right) }\right] \\&+\sum _{\ell =i}^{j}\left[ \left( i-1\right) \alpha _{\ell -i+1}^{\left( 1\right) }x_{j-\ell +1}^{\left( 2\right) }-\left( \ell -1\right) \alpha _{j-\ell +1}^{\left( 1\right) }x_{\ell -i+1}^{\left( 2\right) }\right] \\&+\sum _{\ell =i}^{j}\left[ \left( i-1\right) \left( \ell -1\right) \left\{ \alpha _{\ell -i+1}^{\left( 1\right) }\alpha _{j-\ell +1}^{\left( 2\right) }-\alpha _{j-\ell +1}^{\left( 1\right) }\alpha _{\ell -i+1}^{\left( 2\right) }\right\} \right] \\&=\sum _{\ell =1}^{j-i+1}\left[ \left( \ell -1\right) \alpha _{j-i-\ell +2}^{\left( 2\right) }x_{\ell }^{\left( 1\right) }-\left( j-i-\ell +1\right) \alpha _{\ell }^{\left( 1\right) }x_{j-i-\ell +2}^{\left( 2\right) }\right] \\&+\left( i-1\right) \sum _{\ell =1}^{j-i+1}\left( i-j+2\ell -2\right) \alpha _{\ell }^{\left( 1\right) }\alpha _{j-i-\ell +2}^{\left( 2\right) }. \end{aligned}$$

Indeed, if we put $i=1$, we obtain

$$\begin{aligned} z_{1j}=\sum _{\ell =1}^{j}\left[ \left( \ell -1\right) \alpha _{j-\ell +1}^{\left( 2\right) }x_{\ell }^{\left( 1\right) }-\left( j-\ell \right) \alpha _{\ell }^{\left( 1\right) }x_{j-\ell +1}^{\left( 2\right) }\right] . \end{aligned}$$

The following statement is also easy to check.

Lemma 14

For $1\le i,j\le n-1$ with $i\le j$,

$$\begin{aligned} z_{\left( i+1\right) ,\left( j+1\right) }&=z_{ij}+\sum _{\ell =1}^{j-i+1}\left( i-j+2\ell -2\right) \alpha _{\ell }^{\left( 1\right) }\alpha _{j-i-\ell +2}^{\left( 2\right) }. \end{aligned}$$

(22)

Theorem 15

Consider two matrices $A_{1},A_{2}$ in the McCoy’s form, i.e. $A_{k}=\left[ x_{j-i+1}^{\left( k\right) }+\left( i-1\right) \alpha _{j-i+1}^{\left( k\right) }\right] _{i\le j}$ for $k=1,2$. These matrices commute if and only if there exist some $\xi$ ($1\le \xi \le n$) and r such that

$$\begin{aligned} \alpha _{\sigma }^{\left( 2\right) }&={\left\{ \begin{array}{ll} 0=\alpha _{\sigma }^{\left( 1\right) } &{} \left( \sigma =1,\ldots ,\xi -1\right) \\ r\alpha _{\sigma }^{\left( 1\right) } &{} \left( \sigma =\xi ,\ldots ,n-1\right) , \end{array}\right. } \end{aligned}$$

(23)

where $\alpha _\xi ^{(1)} \ne 0$ (the roles of $A_1$ and $A_2$ shold be changed if $\alpha _\xi ^{(1)}=0$ but $\alpha _\xi ^{(2)}\ne 0$ for $\xi <n$), and

$$\begin{aligned} x_{\sigma }^{\left( 2\right) }=rx_{\sigma }^{\left( 1\right) }\left( \sigma =2,\ldots ,n-\xi +1\right) . \end{aligned}$$

(24)

(assuming that $x_\sigma ^{(1)}$ and $x_\sigma ^{(2)}$ are arbitrary for all $\sigma =1,\ldots ,n$ if $\xi =n$).

Proof

It is easy to check directly from Lemma 13 that these matrices commute when (23) and (24) are satisfied. We show the converse in the following steps. Suppose $A_{1}$ and $A_{2}$ commute.

Step 1:

Assume that $\alpha _{1}^{\left( 1\right) }\ne 0$. Then, one can inductively show that $\alpha _{\sigma }^{\left( 2\right) }=r\alpha _{\sigma }^{\left( 1\right) }$ for all $\sigma$, where $r=\left( \alpha _{1}^{\left( 2\right) }/\alpha _{1}^{\left( 1\right) }\right)$. Firstly, $\alpha _{1}^{\left( 2\right) }=\left( \alpha _{1}^{\left( 2\right) }/\alpha _{1}^{\left( 1\right) }\right) \alpha _{1}^{\left( 1\right) }$ is obvious. Secondly, assume that $\alpha _{\sigma }^{\left( 2\right) }=r\alpha _{\sigma }^{\left( 1\right) }$ for $\sigma =1,\ldots ,N-1$ ($2\le N\le n-1$). Using (22),

$$\begin{aligned} z_{2,\left( N+1\right) }&=\sum _{\ell =1}^{N}\left( -N+2\ell -1\right) \alpha _{\ell }^{\left( 1\right) }\alpha _{N-\ell +1}^{\left( 2\right) }\\&=\left( 1-N\right) \alpha _{1}^{\left( 1\right) }\alpha _{N}^{\left( 2\right) }+\left( N-1\right) \alpha _{N}^{\left( 1\right) }\alpha _{1}^{\left( 2\right) }\\&+\left\{ \left( 3-N\right) \left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{2}^{\left( 1\right) }\alpha _{N-1}^{\left( 1\right) }+\left( N-3\right) \left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{N-1}^{\left( 1\right) }\alpha _{2}^{\left( 1\right) }\right\} \\&+\cdots \\&+\left\{ -\left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{N/2}^{\left( 1\right) }\alpha _{N/2+1}^{\left( 1\right) }+\left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{N/2}^{\left( 1\right) }\alpha _{N/2+1}^{\left( 1\right) }\right\} \\&=\left( 1-N\right) \alpha _{1}^{\left( 1\right) }\alpha _{N}^{\left( 2\right) }+\left( N-1\right) \alpha _{N}^{\left( 1\right) }\alpha _{1}^{\left( 2\right) } \end{aligned}$$

if N is even, and

$$\begin{aligned} z_{2,\left( N+1\right) }&=\sum _{\ell =1}^{N}\left( -N+2\ell -1\right) \alpha _{\ell }^{\left( 1\right) }\alpha _{N-\ell +1}^{\left( 2\right) }\\&=\left( 1-N\right) \alpha _{1}^{\left( 1\right) }\alpha _{N}^{\left( 2\right) }+\left( N-1\right) \alpha _{N}^{\left( 1\right) }\alpha _{1}^{\left( 2\right) }\\&+\left\{ \left( 3-N\right) \left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{2}^{\left( 1\right) }\alpha _{N-1}^{\left( 1\right) }+\left( N-3\right) \left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{N-1}^{\left( 1\right) }\alpha _{2}^{\left( 1\right) }\right\} \\&+\cdots \\&+\left\{ -2\left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{\left\lfloor N/2\right\rfloor }^{\left( 1\right) }\alpha _{\left( 3N-3\right) /2}^{\left( 1\right) }+2\left( \frac{\alpha _{1}^{\left( 2\right) }}{\alpha _{1}^{\left( 2\right) }}\right) \alpha _{\left( 3N-3\right) /2}^{\left( 1\right) }\alpha _{\left\lfloor N/2\right\rfloor }^{\left( 1\right) }\right\} \\&=\left( 1-N\right) \alpha _{1}^{\left( 1\right) }\alpha _{N}^{\left( 2\right) }+\left( N-1\right) \alpha _{N}^{\left( 1\right) }\alpha _{1}^{\left( 2\right) } \end{aligned}$$

if N is odd. Since $N\ne 1$ and $\alpha _{1}^{\left( 1\right) }\ne 0$, we obtain $\alpha _{\sigma }^{\left( 2\right) }=r\alpha _{\ell }^{\left( 1\right) }$ for all $\sigma$ by induction. Using this statement, it is easy to show from (21) that $x_{\sigma }^{\left( 2\right) }=rx_{\sigma }^{\left( 1\right) }$ for all $\ell$.

Step 2:

Assume that $\alpha _{\sigma }^{\left( 1\right) }=0$ for $\sigma =1,\ldots ,\xi -1$ (here we assume $2\le \xi \le n-1$) and suppose $\alpha _{\xi }^{\left( 1\right) }\ne 0$. Then, $\alpha _{\sigma }^{\left( 2\right) }=0$ ($\sigma =1,\ldots ,\xi -1$), which can be derived inductively from (22):

$$\begin{aligned} z_{2,\left( \xi +1\right) }&=\left( \xi -1\right) \alpha _{\xi }^{\left( 1\right) }\alpha _{1}^{\left( 2\right) }=0\\ z_{2,\left( \xi +2\right) }&=\left( \xi -2\right) \alpha _{\xi }^{\left( 1\right) }\alpha _{2}^{\left( 2\right) }+\xi \alpha _{\xi +1}^{\left( 1\right) }\alpha _{1}^{\left( 2\right) }=0\\&\cdots \\ z_{2,\left( \xi +m\right) }&=\left( \xi -m\right) \alpha _{\xi }^{\left( 1\right) }\alpha _{m}^{\left( 2\right) } \\& +\sum _{\ell =1}^{m-1}\left( 2\ell +\xi -m\right) \alpha _{\xi +\ell }^{\left( 1\right) }\alpha _{m-\ell }^{\left( 2\right) }=0\left( m\in \mathbb {N}\right) \\&\cdots \\ z_{2,n}&=\left( 2\xi -n\right) \alpha _{\xi }^{\left( 1\right) }\alpha _{n-\xi }^{\left( 2\right) }+\sum _{\ell =\xi +1}^{n-1}\left( 2\ell -n\right) \alpha _{\ell }^{\left( 1\right) }\alpha _{n-\ell }^{\left( 2\right) }=0 \end{aligned}$$

In addition, one can inductively show that $\alpha _{\sigma }^{\left( 2\right) }=r\alpha _{\sigma }^{\left( 1\right) }$ for all $\sigma =\xi ,\ldots ,n-1$, where $r=\left( \alpha _{\xi }^{\left( 2\right) }/\alpha _{\xi }^{\left( 1\right) }\right)$. Firstly, $\alpha _{\xi }^{\left( 2\right) }=\left( \alpha _{\xi }^{\left( 2\right) }/\alpha _{\xi }^{\left( 1\right) }\right) \alpha _{\xi }^{\left( 1\right) }$ is obvious. Secondly, assume that the statement holds for $\sigma =\xi ,\ldots ,\xi +m$ for some $m\in \left\{ 0,\ldots ,n-\xi -1\right\}$. Then, by (22):

$$\begin{aligned} z_{2,\left( 2\xi +m+1\right) }&=\sum _{\ell =\xi }^{\xi +m+1}\left( 2\ell -2\xi -m-1\right) \alpha _{\ell }^{\left( 1\right) }\alpha _{2\xi +m-\ell +1}^{\left( 2\right) }\\&=\left( -m-1\right) \alpha _{\xi }^{\left( 1\right) }\alpha _{\xi +m+1}^{\left( 2\right) }+\left( m+1\right) \alpha _{\xi +m+1}^{\left( 1\right) }\alpha _{\xi }^{\left( 2\right) }\\&+\left( \alpha _{\xi }^{\left( 2\right) }/\alpha _{\xi }^{\left( 1\right) }\right) \sum _{\ell =1}^{m}\left( 2\ell -m-1\right) \alpha _{\ell +\xi }^{\left( 1\right) }\alpha _{\xi +m-\ell +1}^{\left( 1\right) }\\&=\left( -m-1\right) \alpha _{\xi }^{\left( 1\right) }\alpha _{\xi +m+1}^{\left( 2\right) }+\left( m+1\right) \alpha _{\xi +m+1}^{\left( 1\right) }\alpha _{\xi }^{\left( 2\right) }, \end{aligned}$$

where

$$\begin{aligned}&\sum _{\ell =1}^{m}\left( 2\ell -m-1\right) \alpha _{\ell +\xi }^{\left( 1\right) }\alpha _{\xi +m-\ell +1}^{\left( 1\right) } \\&=\left\{ \left( 1-m\right) \alpha _{\xi +1}^{\left( 1\right) }\alpha _{\xi +m}^{\left( 1\right) }+\left( m-1\right) \alpha _{\xi +m}^{\left( 1\right) }\alpha _{\xi +1}^{\left( 1\right) }\right\} \\&+\cdots \\&+\left\{ -\alpha _{\xi +m/2}^{\left( 1\right) }\alpha _{\xi +m/2+1}^{\left( 1\right) }+\alpha _{\xi +m/2+1}^{\left( 1\right) }\alpha _{\xi +m/2}^{\left( 1\right) }\right\} \\&=0 \end{aligned}$$

if m is even and

$$\begin{aligned}&\sum _{\ell =1}^{m}\left( 2\ell -m-1\right) \alpha _{\ell +\xi }^{\left( 1\right) }\alpha _{\xi +m-\ell +1}^{\left( 1\right) }\\&=\left\{ \left( 1-m\right) \alpha _{\xi +1}^{\left( 1\right) }\alpha _{\xi +m}^{\left( 1\right) }+\left( m-1\right) \alpha _{\xi +m}^{\left( 1\right) }\alpha _{\xi +1}^{\left( 1\right) }\right\} \\&+\cdots \\&+\left\{ -2\alpha _{\xi +m/2}^{\left( 1\right) }\alpha _{\xi +m/2+1}^{\left( 1\right) }+2\alpha _{\xi +m/2+1}^{\left( 1\right) }\alpha _{\xi +m/2}^{\left( 1\right) }\right\} +0\\&=0 \end{aligned}$$

if m is odd. Hence, we obtain $\alpha _{\xi +m+1}^{\left( 2\right) }=r\alpha _{\xi +m+1}^{\left( 1\right) }$, which means that the statement is true for all $\sigma$. Using this statement and (21) for $j=\xi +1,\ldots n$, it is easy to show that $x_{\sigma }^{\left( 2\right) }=rx_{\sigma }^{\left( 1\right) }$ for all $\sigma =2,\ldots ,n-\xi +1$, where $r=\left( \alpha _{\xi }^{\left( 2\right) }/\alpha _{\xi }^{\left( 1\right) }\right)$.

Step 3:

Finally, suppose $\alpha _\sigma ^{(1)}=\alpha _\sigma ^{(2)}=0$ for all $\sigma$. Then, one can easily check from Lemma 13 that the matrices $X_1$ and $X_2$ commute.

Using the statement above, Lemma 11 follows straightforward.