I am wondering how I can show that the solution ($P$) to the following matrix differential equation

$$\begin{align} \dot{P}(t) &= -A^{\top}P - PA + PBR^{-1}B^{\top}P - Q \\[3mm] P(t_{1}) &= S \end{align}$$


$$\begin{align} A = \begin{bmatrix}-4&0\\0&0\end{bmatrix},\quad B = \begin{bmatrix}2\\-2\end{bmatrix},\quad Q = \begin{bmatrix}4&4\\4&4\end{bmatrix},\quad S = \begin{bmatrix}0&0\\0&0\end{bmatrix}\quad \text{and}\quad R= \begin{bmatrix}1&0\\0&1\end{bmatrix}, \end{align}$$

is positive definite (i.e., $P > 0$) without solving the matrix differential equation. Is there some way we can see this, maybe by first rewriting the RHS? What are some useful properties one could use?

For those of you unfamiliar with this equation, there actually exists a unique solution $P$ to this equation if $Q$ and $S$ are positive semidefinite and $R$ positive definite, which they all are in case. It is also true that the solution $P$, in general, is positive semidefinite. I do not have the proof for this, unfortunately.

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    $\begingroup$ What time interval are you considering? Because obviously $P(t_1)$ isn't positive definite, but also for $\tau=t_1+\delta$, with $\delta$ some small positive real number, $P(\tau)$ is approximately equal to $-\delta\,Q$ which would even be negative semi-definite. $\endgroup$ Dec 18, 2020 at 19:37
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    $\begingroup$ @KwinvanderVeen Yes, you are right. The time interval of interest is $t\in[0,t_{1})$. I have found a somewhat satisfactory proof of this, I will present it soon. How do you reach that conclusion, are you making a Taylor expansion to obtain that result? $\endgroup$ Dec 18, 2020 at 20:20
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    $\begingroup$ yes, Taylor expansion using $P(t_1)=0$ and $\dot{P}(t_1)=-Q$. $\endgroup$ Dec 18, 2020 at 21:14


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