Consider the linear matrix differential equation

$\def\diag{\mathrm{diag}}$ \begin{align} U(0) &= I\\ \frac{\mathrm{d}U}{\mathrm{d}t}(t) &= U(t) \phantom{.} Q(t) & & \quad(1) \end{align}

where $Q(t),U(t)$ are $n\times n$ real valued matrices and $Q(t)$ is a transition rate matrix, which means that the off diagonal entries are nonnegative and each row sums to zero. Unfortunately, in general $Q(t_1)Q(t_2)\neq Q(t_2)Q(t_1)$ so the tempting equality $U(t)=\exp\left(\int_0^t Q(s)\,\mathrm{d}s\right)$ is false in general.

For some $\delta>0$, consider a "magnified" process $V^\delta$

\begin{align} V^\delta(0) &= I\\ \frac{\partial V^\delta}{\partial t}(t) &= V^\delta(t) \phantom{.} (1+\delta)Q(t) & & \quad(2) \end{align}

Suppose one can compute the solution of (1) explicitly. Is there a simple expression for the solution of (2) in terms of the solution of (1)?

Actually I am only interested in calculating $\left. \frac{\partial V^\delta (t)}{\partial \delta}\right|_{\delta=0}$, which may be easier.



1 Answer 1


Let's suppose all the $V^\delta(t)$ have a common initial condition $V^\delta(0) = U_0$. If $W^\delta(t) = \dfrac{\partial}{\partial \delta} V^\delta(t)$, then we have $$ \dfrac{dW^\delta}{dt} = \dfrac{\partial}{\partial \delta} \dfrac{dV^\delta}{dt} = \dfrac{\partial}{\partial \delta} (1+\delta) V^\delta(t) Q(t) = V^\delta(t) Q(t) + (1+\delta) W^\delta(t) Q(t)$$ with $W^\delta(0) = 0$. In particular, for $\delta = 0$ we have $$ \dfrac{dW^0}{dt} = (U(t) + W^0(t)) Q(t)$$ The solution to this is $$W^0(t) = Z(t) U(t)$$ where $$ Z(t) = \int_0^t U(s) Q(s) U(s)^{-1}\; ds $$

  • $\begingroup$ Thank you very much! Is the non-commutativity of $Q$ not a problem in computation of $Z$ ? $\endgroup$
    – user50085
    Sep 26, 2016 at 4:53
  • 1
    $\begingroup$ Surprisingly not. It's straightforward to check that this does satisfy the differential equation. $\endgroup$ Sep 26, 2016 at 4:57
  • $\begingroup$ Thanks a lot! I tried to upvote, but not enough credits and tried to write a note, but not enough characters :) $\endgroup$
    – user50085
    Sep 26, 2016 at 21:04

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