Let $P_{m,n}=P_{m,n}(x,y)$ be a polynomial family. Here is some initial terms $$ P_{0,0}=1, P_{1,0}=2x, P_{0,1}=2y, P_{1,1}=8xy.$$ I know that the polynomials for any $m,n \geq 0$ satisfy the five differential recurrence relations \begin{align} &n \frac{\partial P_{m,n-1}}{\partial x}=m \frac{\partial P_{m-1,n}}{\partial y},\\ & x \frac{\partial P_{m,n}}{\partial x}=m P_{m,n}+m\frac{\partial P_{m-1,n}}{\partial x},\\ & y\frac{\partial P_{m,n}}{\partial x}=m P_{m-1,n+1}+n \,\frac{\partial P_{m,n-1}}{\partial x},\\ & y \frac{\partial P_{m,n}}{\partial y}=n P_{m,n}+n\frac{\partial P_{m,n-1}}{\partial y},\\ & x\frac{\partial P_{m,n}}{\partial y}=n P_{m+1,n-1}+m \,\frac{\partial P_{m-1,n}}{\partial y}. \end{align}
Also. they satisfied the differential equation $$ (1-x^2) \frac{\partial^2 P_{m,n}}{\partial x^2} -x y \frac{\partial^2 P_{m,n} }{\partial x \partial y} -(n+3) x \frac{\partial P_{m,n}}{\partial x }+m y \frac{\partial P_{m,n}}{\partial y }+m(m+n+2) P_{m,n}=0, $$ for any $m,n.$
I need to eliminate all the derivatives and get pure recurrence relations for $P_{m,n}$.
By numeric expеriments I guess such recurrence relations $$ 2 (1{+}m{+}n) x P_{m,n}=P_{m+1,n}{-}n(n{-}1)P_{m+1,n-2}{+m(m+2n+1)}P_{m-1,n},\\ 2 (1{+}m{+}n) y P_{m,n}=P_{m,n+1}{-}m(m{-}1)P_{m-2,n+1}{+}n(n+2m+1)P_{m,n-1} $$ but I still cant prove it.
Any help?
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