Recurrence Relation with multiple variables

Question

Consider the following recurrence relation with two variables and similar bounding conditions:

\begin{align*} x_{1,t} &= \frac{1}{2-\frac{2}{n}} x_{1,t-1} + \frac{1}{2} x_{2,t-1} \\ x_{i,t} &= \frac{1}{2} x_{i-1,t-1} + \frac{1}{2} x_{i+1,t-1} & 1 < i < n \\ x_{n,t} &= \frac{1}{2} x_{n-1,t-1} + \frac{1}{2-\frac{2}{n}} x_{n,t-1} \end{align*}

$x_{i,0} = 1 $ for all $1 \leq i \leq n$. Is there any standard method to solve these?

Answer 1

Your system reads as $$ {\bf x}(t) = {1 \over 2}\;\left( {\matrix{ {1 + {1 \over {n - 1}}} & 1 & 0 & \cdots & 0 \cr 1 & 0 & 1 & \cdots & 0 \cr \vdots & \ddots & \ddots & \ddots & \vdots \cr 0 & \cdots & 1 & 0 & 1 \cr 0 & \cdots & 0 & 1 & {1 + {1 \over {n - 1}}} \cr } } \right)\;{\bf x}(t - 1) $$

So it is a random walk, with a stop at the ends, and with the immission (always at the ends) of a influx given as constant percentage increase of the population therein.
And the increase percentage lowers at growing of the vector dimension.

Since the matrix diagonalization is not neat, you are probably looking for an approximation. If so with respect to what ?

-- reply to your comment --

The matrix above (call it $\bf M$) is symmetric, either wrt the diagonal and wrt the antidiagonal , i.e. $$ {\bf M} = {\bf M}^{\,T} = {\bf J}\;{\bf M}\;{\bf J} $$ where ${\bf J}$ is the exchange matrix.
Thus the matrix is centrosymmetric and in particular symmetric tridiagonal, and only almost-Toeplitz symmetric.

As such:
- $\bf M$ can be diagonalized by an orthogonal matrix $\bf Q$;
- all its eigenvalues are real and distinct;
- if $\bf v$ is an eigenvector so is its symmetric ${\bf J}{\bf v}$;
- since $Q$ is orthogonal, it will contain the eigenvectors.

For an eigenvector we surely have that the ratio of the components is constant, and v.v. a vector for which the ratio remains constant must be an eigenvector.
A calculation for small $n$ reveals that the e.v. associated to the higher eigenvalue has all the components positive and is in fact symmetric.
But I couldn't yet reach to a demonstration of that, nor of a bound in such a ratio.