The solution of $dX(t)/dt = A X(t) + X(t) A + Q$?

Question

According to the perfect answer from A.Γ. in Any compact solution for $dX/dt = A X(t) + X(t) A^T$?, I understand that the solution of

$$\frac{dX(t)}{dt} = AX(t) + X(t)A^T,$$

where $A, X(t) \in {\mathbb R}^{n\times n}$ ($X(t) = X^T(t)$), is

$$X(t) = e^{At}X(0)e^{A^Tt}.$$

Now, consider the following ode

$$\frac{dX(t)}{dt} = AX(t) + X(t)A^T + Q,$$

where $Q$ is a symmetric matrix in ${\mathbb R}^{n\times n}$. I guess the solution is

$$X(t) = e^{At}X(0)e^{A^Tt} + \int_0^t e^{A(t-\tau)}Qe^{A^T(t-\tau)}d\tau.$$

The reason is when I take its derivative, I can get ${dX(t)}/{dt} = AX(t) + X(t)A^T + Q$. Am I right?

Answer 1

A rigorous way.

$$ X'=AX+XA^T+Q \quad\Longrightarrow\quad e^{-tA}(X'-AX)=e^{-tA}XA^T+e^{-tA}Q \quad\Longrightarrow\quad (e^{-tA}X)'=e^{-tA}XA^T+e^{-tA}Q \quad\Longrightarrow\quad \big((e^{-tA}X)'-e^{-tA}XA^T\big)e^{-tA^T}=e^{-tA}Q e^{-tA^T}\quad\Longrightarrow\quad \big(e^{-tA}Xe^{-tA^T}\big)'=e^{-tA}Q e^{-tA^T}\quad\Longrightarrow\quad e^{-tA}X(t)e^{-tA^T}=X(0)+\int_0^te^{-sA}Q e^{-sA^T}ds\quad\Longrightarrow\quad X(t)=e^{tA^T}X(0)e^{tA^T}+\int_0^te^{(t-s)A}Q e^{(t-s)A^T}ds. $$

Answer 2

Follow the same idea as in the previous question. Define, $$ Y(t) = \int_0^t e^{-A\tau}Qe^{-A^T\tau}d\tau,\\ \Phi(t) = e^{-At}X(t)e^{-A^Tt} - Y(t) . $$ Next compute the derivatives, $$ \dot{Y} = e^{-At}Qe^{-A^Tt},\\ \dot{\Phi} = e^{-At}[-A X + \dot{X} - XA^T]e^{-A^Tt} -e^{-At}Qe^{-A^Tt},\\ =e^{-At}[\underbrace{-A X + \dot{X} - XA^T - Q}_{=0}]e^{-A^Tt}=0 $$ by using the differential equation for $X$. As before, knowing that $\Phi$ is a constant allows one to pick a value, say $t=0$, to write, $$ \Phi(t) = \Phi(0) =X(0) - Y(0)=X(0),\\ \Rightarrow X(t) = e^{At}\left[X(0) + \int_0^t e^{-A\tau}Q e^{-A^T\tau} d\tau \right]e^{A^Tt} $$ by re-arranging. This matches the suggested solution.