1
$\begingroup$

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 $\lambda$.

Improve this question
$\endgroup$
5
  • $\begingroup$ If you put $b_i=f(i,N-3-i)$ then you get recurrence relations with one indices $\endgroup$ Commented Dec 12, 2023 at 12:52
  • $\begingroup$ Thanks for your comment! But how then do I deal with terms $f(i,j)$ where $j \neq N-3-i$? $\endgroup$ Commented Dec 12, 2023 at 13:00
  • $\begingroup$ I just corrected a mistake in the first equation by the way... It should hold for $i+j < N-3$ instead of $i+j = N-3$. $\endgroup$ Commented Dec 12, 2023 at 13:01
  • $\begingroup$ I want to find an explicit expression for $f(i,j)$ in terms of $i$, $j$, $\lambda$ and $N$. Should have specified that indeed... $\endgroup$ Commented Dec 12, 2023 at 17:20
  • $\begingroup$ This looks like the discrete Laplace equation on the square lattice, with a certain boundary condition, see, e.g., en.wikipedia.org/wiki/Discrete_Laplace_operator . $\endgroup$ Commented Dec 12, 2023 at 19:49

2 Answers 2

Reset to default
3
$\begingroup$

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}.

Improve this answer
$\endgroup$
11
  • $\begingroup$ I get one solution for $n=6$ with $f(0,0)=\frac{48r+11}{62208r^5+77760r^4+34560r^3+6660r^2+525r+11}$. Sage $\endgroup$ Commented Dec 13, 2023 at 8:52
  • $\begingroup$ Thanks a lot for your elaborate response! The one thing that is missing in my system of equations I think is the fact that $f$ is symmetrical, that is $f(i,j) = f(j,i)$ for each $i,j$. So in case of $n = 6$, if we only consider $f(i,j)$ with $i \geq j$, then we are left with only 9 variables instead of 15 and then the system would have only one solution right? Do you think a closed form solution would exist in this case? $\endgroup$ Commented Dec 13, 2023 at 9:42
  • $\begingroup$ Also, the Markov chain/random walk interpretation is in fact where I got this system from! What I'm actually trying to find is the value of $\mathbb{E}[\lambda^{\tau_{i,j}]}$, where $\tau_{i,j}$ is the hitting time to this sloped boundary of the discrete triangle $T$ from a point $(i,j)$. But the chain is a little bit different from what you ar suggesting: from each point you can jump to a neighbouring point with probability $1/{2n}$ and you stay at the same point with the remaining probability (which is $1-4/{2n}$ in case of 4 neighbours and $1-3/{2n}$ in case of 3 neighbours etc.) $\endgroup$ Commented Dec 13, 2023 at 9:49
  • $\begingroup$ Ah, I've found the discrepancy. I misread the question as what @user675763 surely intended, which is to replace all of the $< N - 3$ with $\le N - 3$ so that every cell in the triangle has a constraint, rather than having an antidiagonal at $i + j = N - 3$ which is unconstrained. $\endgroup$ Commented Dec 13, 2023 at 12:18
  • $\begingroup$ Ah yes!! You are completely right! Thanks for pointing this out @PeterTaylor! $\endgroup$ Commented Dec 13, 2023 at 13:00
1
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, I agree it looks quite hopeless... $\endgroup$ Commented Dec 15, 2023 at 16:23
  • $\begingroup$ But thanks for giving it some thought anyway! :) $\endgroup$ Commented Dec 15, 2023 at 16:25

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.