# Is it possible to express the solution of a matrix Riccati differential equation as an eigenvalue problem?

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

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

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.

• 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? Dec 28, 2022 at 9:59
• $\hat{X}$ 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. Dec 28, 2022 at 14:53