Solving a very particular 2 variable recursion relation

Question

While solving a problem, me and a friend hit upon a particular recurrence relation that we tried solving using generating functions but failed. The relation looks like this :

$$P(n,k)=pP(n-1,k+1)+qP(n-1,k-1)$$ with the conditions

$$n\in\mathbb N\cup \{0\},\ 0\le k\le m$$

$$P(n,0)=0,\ P(0,k)=\begin{cases}0\ \ \text{if $k<m$}\\ 1\ \ \text{if $k=m$} \end{cases}$$

$$P(n,k)=0, \text{when $k>m$ or $n<0$}$$

where $m\in\mathbb N$ is a given value.

Here's what we tried. Let us look at a generating function $P(x,y)=\sum_{n,k}P(m,n)x^ny^k$ that models the recursion. Then, we note that:

\begin{multline}P(x,y)(y-px-qxy^2)=\sum_{n,k}P(n,k)(x^ny^{k+1}-px^{n+1}y^k-qx^{n+1}y^{k+2})\\ \hspace{10em}=\sum_{\substack{n\ge 1\\ k\ge 2}}P(n,k-1)-pP(n-1,k)-qP(n-1,k-2)\\ \hspace{7em}+\sum_{n=0}^\infty P(n,1)(x^ny^2-px^{n+1}y-qx^{n+1}y^3)\hspace{2em}+(y^{m+1}-pxy^m-qxy^{m+2})\end{multline}

The first summation vanishes dues to the recurrence relation, and thus we get:

\begin{multline}P(x,y)(y-px-qxy^2)=\sum_{n=0}^\infty P(n,1)(x^ny^2-px^{n+1}y-qx^{n+1}y^3)\\+(y^{m+1}-pxy^m-qxy^{m+2})\end{multline}

But after this, we are stuck.

Any help with solving this is greatly appreciated.

Edit : As lonza leggiera points out, the boundary conditions are contradictory to the fact that $p,q\ne 0$. We found out where we went wrong. The correct recursion should be:

$$P(n,k)=pP(n-1,k+1)+qP(n-1,k-1)$$ with the conditions

$$p+q=1$$ $$P(n,k)=0,\ \text{if $n<0$ or $k\le 0$}$$ $$P(0,k)=0,\ \text{if $k< m$}$$ $$P(n,k)=1,\ \text{if $k\ge m$}$$

Answer 1

This is not really an answer, but a demonstration that your currently specified conditions are probably not quite what you want. First, the specifications

with the conditions $$ n\color{red}{\in\mathbb{N}\cup\{0\}}, 0\le k\color{red}{\le m} $$

and

$$ P(n,k)=0,\ \text{when }\ \color{red}{k>m}\ \text{or}\ \color{red}{n<0} $$

are contradictory. If you ignore the red parts of the first set of conditions and impose just the second set, then I believe a solution will only exist if $\ p=q=0\ $, when the (unique) solution is $\ P(n,k)=0\ $. From the condition $$ P(n,k)=0\ \ \text{when }\ k>m\ , $$ for example, we get $$ P(0,m+2)=P(0,m+1)=P(1,m+1)=0 $$ The recursion relation then gives us $$ P(1,m+1)=pP(0,m+2)+qP(0,m)=q\ , $$ from which it follows that $\ q=0\ $. Showing that $\ p=0\ $ requires a little more work, which I won't bother doing here.

If you instead limit the domain of the solution you're looking for to be integers satisfying $$ n\ge0, 0\le k\le m+1\ , $$ and simply require $ P(n,m+1)=P(n,0)=0\ $ for all $\ n\in\mathbb{N}\cup\{0\}\ $, then I believe there will always be a non-trivial unique solution for which you may be able to obtain a fairly simple expression in closed form. I'm not inclined to bother trying to find it, however, unless you can say definitely that it really is the solution you're looking for.