# Evaluate derivative of reachability Gramian: is this correct?

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?

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.

• Ah yes, well then, one less mystery! Thank you. Jan 4, 2021 at 22:27
• @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 = Jan 4, 2021 at 22:31
• Great! I'll remember that. 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}