This is related to my previous question How can I express the solution to a discrete Lyapunov equation as an eigenvalue problem?.

Given the algebraic Riccati equation (ARE) $$A^T X + XA + XRX + Q = 0$$ The solution $X \in \mathbb R^{n \times n}$ that satisfies this equation is $X = X_2X_1^{-1}$, where $\begin{bmatrix} X_1 \\ X_2\end{bmatrix}$ is a $2n \times n$ matrix whose columns are the $n$ eigenvectors associated with the $n$ stable eigenvalues of the Hamiltonian matrix $$H = \begin{bmatrix} A & R \\ -Q & -A^T\end{bmatrix}$$ I am now interested in the solution of the differential Riccati equation (DRE) $$\dot X(t) = -X(t)A - A^TX(t) - X(t) R X(t) + Q$$ subject to the initial condition $X(0) = X_0$. In particular, I'm aware that the solution $X(t)$ can be expressed as $X(t) = X_2(t) X_1^{-1}(t)$, where $X_1(t)$ and $X_2(t)$ satisfy the linear system of differential equations $$\begin{bmatrix}\dot{X}_1(t) \\ \dot X_2(t)\end{bmatrix} = H \begin{bmatrix}X_1(t) \\ X_2(t)\end{bmatrix}$$ subject to the initial condition $$\begin{bmatrix}X_1(0) \\ X_2(0)\end{bmatrix} = \begin{bmatrix} I \\ X_0 \end{bmatrix}$$ where the Hamiltonian matrix $H$ is the one defined above.

The solution $X(t)$ to the DRE and the solution $X$ to the ARE look very similar, yet I cannot solve the DRE in a similar way. Here is how I tried to solve the DRE, which is the same approach that I would use to solve the ARE. First, I would rewrite the DRE as follows: \begin{align} \dot X(t) &= -X(t)A - A^TX(t) - X(t) R X(t) + Q \\ \dot X(t) &= \begin{bmatrix} X(t) & -I\end{bmatrix} \begin{bmatrix} A + RX(t) \\ -Q + A^T X(t) \end{bmatrix} \\ \dot X(t) &= \begin{bmatrix} X(t) & -I\end{bmatrix} \begin{bmatrix} A & R \\ -Q & A^T\end{bmatrix} \begin{bmatrix} I \\ X(t)\end{bmatrix} \end{align} However, I'm not sure how to proceed from here. Normally, the left-hand side of this equation would be the $0$ matrix instead of $\dot X(t)$, from which I can proceed. That is not the case here.

What I’m looking for is a way to derive the solution $X(t) = X_2(t) X_1^{-1}(t)$ to the DRE using an approach similar to the one used to derive the solution $X = X_2 X_1^{-1}$ to the ARE

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    $\begingroup$ Since you pinged me in another thread: sorry but I don't know a way to do it; I suspect the answer is no. $\endgroup$ Commented Dec 18, 2022 at 8:22
  • $\begingroup$ @FedericoPoloni thank you for taking the time to read through this question! $\endgroup$
    – mhdadk
    Commented Dec 18, 2022 at 8:26

1 Answer 1


You can cast the DRE to an ARE, by changing your point of view a bit.

If $X(t)$ solves the DRE

$$ \dot{X}(t) = A^T X(t) + X(t) A - X(t)RX(t) + Q, $$

then $X(t)$ solves the ARE

$$ 0 = A^T \hat{X} + \hat{X} A - \hat{X}R\hat{X} + \underbrace{Q - \dot{X}(t)}_{\tilde{Q}};$$

the sought after solution is here $\hat{X}$.

Of course, this does not help with the numerical solution but enables you to apply theoretical results valid for the ARE to the DRE.

  • $\begingroup$ How did you go from the DRE to the ARE? As in, where did $\hat X$ come from? Also, isn't $\tilde Q$ time-varying? If so, then is this still an ARE? $\endgroup$
    – mhdadk
    Commented Dec 28, 2022 at 9:59
  • $\begingroup$ $\hat{X}$ is just the unknown of the ARE and $X(t)$ is the corresponding solution. $\tilde{Q}$ is time-dependent and the solution $X(t)$ as well. You can think of $t$ as a parameter. For each instance of $t$ you have a new ARE as the matrix $\tilde{Q}$ changes and $X(t)$ is its solution. $\endgroup$
    – DerZwirbel
    Commented Dec 28, 2022 at 14:53

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