Recurrence relation with two variables

Question

I am stuck on a recurrence relation with two variables. I'm familiar with techniques to solve recurrence relations with one variable and looked into ways to solve recurrence relations with multiple variables, but I somehow always end up with a huge mess. This is the recurrence relation:

\begin{align} [2N(1-\lambda) + 4\lambda]f(i,j) &= \lambda f(i-1,j) + \lambda f(i,j-1) + \lambda f(i+1, j) + \lambda f(i, j+1), \quad i, j > 0 \text{ and } i+j \leq N-3 \end{align} \begin{align} [2N(1-\lambda) + 3\lambda]f(i,0) &= \lambda f(i-1,0) + \lambda f(i,1) + \lambda f(i+1,0), \quad 0 < i \leq N-3 \end{align} \begin{align} [2N(1-\lambda) + 3\lambda]f(0,j) &= \lambda f(0,j-1) + \lambda f(1,j) + \lambda f(0,j+1), \quad 0 < j \leq N-3 \end{align} \begin{align} [2N(1-\lambda) + 2\lambda]f(0,0) &= \lambda f(1, 0) + \lambda f(0,1) \end{align} \begin{align} f(i,j) = 1, \quad i+j = N-2. \end{align}

Here, $N$ is some positive integer ($N \geq 5$) and $\lambda \in (0,1)$. I'm looking for an explicit expression for $f(i,j)$ in terms of $i$, $j$, $N$ and $\lambda$.

Answer 1

The following is an answer to the original version of the question.

Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$, \begin{equation} \begin{aligned} (2n r+4)f(i,j) =& f(i-1,j) + f(i,j-1) + f(i+1, j) + f(i, j+1) \\ &\text{if}\quad 0<i<i+j<n-3, \end{aligned} \tag{1}\label{1} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(i,0) &= f(i-1,0) + f(i,1) + f(i+1,0) \\ &\text{if}\quad 0<i<n-3, \end{aligned} \tag{2}\label{2} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(0,j) &= f(0,j-1) + f(1,j) + f(0,j+1) \\ &\text{if}\quad 0<j<n-3, \end{aligned} \tag{3}\label{3} \end{equation} \begin{equation} \begin{aligned} (2n r+2)f(0,0) &= f(1, 0) + f(0,1), \end{aligned} \tag{4}\label{4} \end{equation} \begin{equation} \begin{aligned} f(i,n-2-i) = 1 \quad\text{if}\quad 0\le i\le n-2, \end{aligned} \tag{5}\label{5} \end{equation} where $n:=N$ and $r:=(1-\lambda)/\lambda\in(0,\infty)$. Note that, if $n=5$, then condition \eqref{1} is vacuous.

Already for $n=6$, there are infinitely many solutions of the linear system \eqref{1}--\eqref{5}. These solutions are given by the following: \begin{equation} \begin{aligned} f(0,4)&=1\\ f(1,0)&=(12 r+2) f(0,0)-f(0,1)\\ f(1,1)&=(12 r+3) f(0,1)-f(0,0)-f(0,2)\\ f(1,2)&=(12 r+3) f(0,2)-f(0,1)-f(0,3)\\ f(1,3)&=1\\ f(2,0)&=(12 r+3) f(1,0)-f(0,0)-f(1,1)\\ f(2,1)&=(12 r+4) f(1,1)-f(0,1)-f(1,0)-f(1,2)\\ f(2,2)&=1\\ f(3,0)&=(12 r+3)f(2,0)-f(1,0)-f(2,1)\\ f(3,1)&=1\\ f(4,0)&=1 \end{aligned} \tag{6}\label{6} \end{equation} We see that, of the the $15$ variables $f(i,j)$ (for $n=6$), the four variables $f(0,0),f(0,1),f(0,2),f(0,3)$ are "free", in the sense that one can give $f(0,0),f(0,1),f(0,2),f(0,3)$ any values and then compute the corresponding values of the remaining $15-4=11$ variables according to \eqref{6}.

Answer 2

The solutions look like a mess, so it's not too surprising that you always end up with one. If we follow Iosif Pinelis in dividing all by the last constraint by $\lambda$ and substituting $r = \frac{1-\lambda}\lambda$ then

N=3 (trivial)

\begin{eqnarray*}f(0, 0) &=& \frac{1}{3r+1} \end{eqnarray*}

N=4

\begin{eqnarray*} f(0, 0) &=& \frac{1}{(8r+1)(2r+1)} \\ f(0, 1) = f(1, 0) &=& \frac{4r + 1}{(8r+1)(2r+1)} \\ \end{eqnarray*}

N=5

\begin{eqnarray*} f(0, 0) &=& \frac{20r + 7}{(50r^2 + 20r + 1)(50r^2 + 40r + 7)} \\ f(0, 1) = f(1,0) &=& \frac{(5r + 1)(20r + 7)}{(50r^2 + 20r + 1)(50r^2 + 40r + 7)} \\ f(0, 2) = f(2, 0) &=& \frac{500r^3 + 450r^2 + 115r + 7}{(50r^2 + 20r + 1)(50r^2 + 40r + 7)} \\ f(1, 1) &=& \frac{500r^3 + 400r^2 + 100r + 7}{(50r^2 + 20r + 1)(50r^2 + 40r + 7)} \\ \end{eqnarray*}

N=6

\begin{eqnarray*} f(0, 0) &=& \frac{48r + 11}{(3r + 1)(144r^2 + 36r + 1)(144r^2 + 96r + 11)} \\ f(0, 1) = f(1, 0) &=& \frac{(6r + 1)(48r + 11)}{(3r + 1)(144r^2 + 36r + 1)(144r^2 + 96r + 11)} \\ f(0, 2) = f(2, 0) &=& \frac{1728r^3 + 1224r^2 + 234r + 11}{(3r + 1)(144r^2 + 36r + 1)(144r^2 + 96r + 11)} \\ f(1, 1) &=& \frac{1728r^3 + 1008r^2 + 192r + 11}{(3r + 1)(144r^2 + 36r + 1)(144r^2 + 96r + 11)} \\ f(0, 3) = f(3, 0) &=& \frac{10368r^4 + 10368r^3 + 3312r^2 + 384r + 11}{(3r + 1)(144r^2 + 36r + 1)(144r^2 + 96r + 11)} \\ f(1, 2) = f(2, 1) &=& \frac{10368r^4 + 9504r^3 + 2880r^2 + 336r + 11}{(3r + 1)(144r^2 + 36r + 1)(144r^2 + 96r + 11)} \\ \end{eqnarray*}

It looks reasonable to conjecture $f(0,1) = f(1,0) = (Nr+1)f(0, 0)$, but I'm not optimistic about getting anything further for the general case.