So I have the reachability Gramian matrix for a linear time-invariant system:

\begin{align} W(t_{0},t) = \int_{t_{0}}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}. \end{align}

In this case I have $t_{0}=0$. Let us differentiate this w.r.t. to $t$:

\begin{align} \dot{W}(0,t) =& \frac{d}{dt}\left[\int_{0}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}ds\right] \\ =& \frac{d}{dt}\left[e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}\right] \\ =& Ae^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t} \\ &+e^{At}\left[e^{-As}BB^{\intercal}e^{-A^{\intercal}s}\bigg\vert_{0}^{t}\right]e^{A^{\intercal}t} + e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}A^{\intercal} \\[3mm] =& AW(0,t) + BB^{\intercal} - e^{At}BB^{\intercal}e^{A^{\intercal}t}+W(0,t)A^{\intercal}. \end{align}

So this is the result I obtain. However, in the solution to this problem that I was trying to solve the term $- e^{At}BB^{\intercal}e^{A^{\intercal}t}$ did not appear, or was zero. So I am wondering if I made some error, or if there is some control theory result that make this term vanish?


2 Answers 2


When you applied the product rule, you incorrectly computed the derivative of the integral. By the fundamental theorem of calculus, we have $$ \frac d{dt} \int_{t_0}^t e^{-As}BB^\top e^{-A^\top s}\,ds = e^{-At}BB^\top e^{-A^\top t}. $$ There is no $BB^\top$ term that should appear here.

  • $\begingroup$ Ah yes, well then, one less mystery! Thank you. $\endgroup$ Jan 4, 2021 at 22:27
  • 1
    $\begingroup$ @SimpleProgrammer You're welcome. By the way, I would recommend that you take a quick look at the edited Latex for your series of equations. If you use the align environment, it is generally good practice to start a new line for every = and have & adjacent to your = $\endgroup$ Jan 4, 2021 at 22:31
  • $\begingroup$ Great! I'll remember that. $\endgroup$ Jan 5, 2021 at 11:05

Applying the Leibniz integral rule to your middle term yields

$$ \frac{d}{dt}\left[\int_{0}^{t} e^{-A\,s} B B^{\intercal} e^{-A^{\intercal} s}ds\right] = \int_{0}^{t} \frac{d}{dt}\left[e^{-A\,s} B B^{\intercal} e^{-A^{\intercal} s}\right] ds + e^{-A\,t} B B^{\intercal} e^{-A^{\intercal} t} = e^{-A\,t} B B^{\intercal} e^{-A^{\intercal} t}. $$

You can also apply the Leibniz integral rule directly to initial expression, which yields

\begin{align} \frac{d}{dt}\left[\int_{0}^{t} e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)}ds\right] &= \int_{0}^{t} \frac{d}{dt}\left[e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)}\right] ds + e^{A (t-t)} B B^{\intercal} e^{A^{\intercal} (t-t)}, \\ &= \int_{0}^{t} \left[A\,e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)} + e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)} A^{\intercal}\right] ds + B B^{\intercal}. \end{align}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .