Solving a linear matrix ODE

Question

I have an equation of the form

$$\dot{P}(t) = A^T P(t) + P(t) A + Q$$

I also have an initial condition $P(0)=P_0$

Only the articles on the algebraic Riccati's equation mention equations of this form, and none of the ones about the Riccati's DE's of this form. However, this is the equation I want to find a solution to. Is there a better option than a numerical methods like Euler's or Runge-Kutta's methods?

I haven't been making any progress for a while now.

Answer 1

This is actually an affine equation or linear plus constant term. The solution can be found with the method of variation of constants.

First define the linear operator (acting on matrices) $$ K_A(X) = A^T X + X A $$

So the solution of the homogeneous ODE (with $Q=0$) is therefore $$ X(t) =e^{tK_A}[X_0] $$

The solution of your equation can be found with the method of variation of constants, and it reads

$$ P(t) = e^{t K_A} [P(0)] + \int_0^t dt e^{(t-s)K_A} [Q]. $$

If you are interesting on solving the ODE at finite time, Runge-Kutta fourth order is indeed one of the most famous method.

With the analytic formula above however you can derive several qualitative properties. For example you can find the fixed point of the ODE by solving the related Lyapunov equation.

Added

Just to be clear

$$ e^{tK_A}[X_0] = e^{tA^T} X_0 e^{tA} $$

Answer 2

Assuming $A$ and $Q$ are constant matrices, as your notation seems to indicate, consider the function $$ H(t)=e^{-A^Tt}P(t)e^{-At} $$ and its derivative $$ \dot H(t)=e^{-A^Tt}(-A^TP(t)+\dot P(t)-P(t)A)e^{-At}=e^{-A^Tt}Qe^{-At}. $$ This transforms the ODE problem into a more simple quadrature problem $$ P(t)=e^{A^Tt}H(t)e^{At},~~~ H(t)=P(0)+\int_0^te^{-A^Ts}Qe^{-As}\,ds. $$