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This problem arose when solving a continuous Markov chain exercise from a book I am studying. Given a set of positive $q_i$ with $i \in \mathbb{Z} $, and non-negative $\lambda$ and $\mu$ that add to 1, the solution amounts to solving

$p_{0,i}'(t) = \lambda q_{i-1} p_{0,i-1}(t) - q_i p_{0,i}(t) + \mu q_{i+1} p_{0,i+1}(t) \;\; \text{where} \; \; i \in \mathbb{Z}$

with initial conditions $p_{0,i}(0) = \delta_{0,i}$. How does one go about solving this problem?

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    $\begingroup$ Try to write it in a vector-matrix form, then you should see it immediately. $\endgroup$ May 17, 2011 at 19:19
  • $\begingroup$ @abatkai Unfortunately I am not understanding how this helps. The solution to this difference differential equation is 0-th row of matrix exponential of a tri-banded matrix, $p_{0,i}(t) = (exp(t*Q))_{0,i}$, where $Q$ is zero except for $Q_{i,i+1}=\lambda q_i$, $Q_{i,i}=-q_i$ and $Q_{i,i-1} = \mu q_i$. Then the question gets rephrased into how does compute matrix exponential of this matrix ? I am sure this is a standard topic, so a pointer to a book, or web-article will be sufficient. $\endgroup$
    – Sasha
    May 17, 2011 at 19:32
  • $\begingroup$ confluential: you say: "The solution to this difference differential equation is 0-th row of matrix exponential..." Therefore, why are you asking this question on MO? Why don't you directly ask "How do we compute exp(M) for a tri-banded matrix M" or similar (after first trying to look it up yourself, of course; I'm sure almost every good numerical analysis book will discuss this at length). $\endgroup$
    – Zen Harper
    May 18, 2011 at 7:52
  • $\begingroup$ @(Zen Harper) Because I am seeking a symbolic solution. I was hoping that reformulating of it in terms of difference-differential equation will lead to results faster. After all, I only need to compute $p_{0,i}(t)$ and not $p_{j,i}(t)$. Maybe I failed to make it more explicit, but the matrix $Q$ is infinite dimensional, so this may be a substantial simplification for exponentiation of the entire matrix. $\endgroup$
    – Sasha
    May 18, 2011 at 14:02
  • $\begingroup$ Basically, your problem looks like a spatial finite difference discretization of a heat equation with (spatially) variable coefficients. You should not expect a closed form solution unless $q_i$ is independent of $i$. $\endgroup$ May 18, 2011 at 14:33

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Thanks to Michael Renardy for the analogy with the heat equation. Let's assume $q_i = q$, i.e. independent on $i$. The difference-differential equation can then be solved by using generating function technique. Let $G(z) = \sum_{i \in \mathbb{Z}} z^i p_{0,i}(t)$. Then the equation and initial condition imply that $G(z) = \exp(-q t) \exp \left( q t (\lambda z + \mu/z) \right)$. Extracting series coefficients from here gives: $p_{0,i}(t) = \exp(-q t) \lambda^{i/2} \mu^{-i/2} I_{i}( 2 q t \sqrt{\lambda \mu})$, which is known as Skellam distribution.

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