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?
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.
$p_{0,i}(t) = (exp(t*Q))_{0,i}$, 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$$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. $\endgroup$