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.



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 '16 at 4:53
  • 1
    $\begingroup$ Surprisingly not. It's straightforward to check that this does satisfy the differential equation. $\endgroup$ – Robert Israel Sep 26 '16 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 '16 at 21:04

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.