How to solve an ODE of this form $\dot{P}(t)=A^TP(t)+P(t)A$?

Question

Background

If $\dot{x}(t)=A \,x(t)$, then we know the solution is $x(t)=e^{At}x(0)$.

Question

Now let $\dot{P}(t)=A^TP(t)+P(t)A$, what is $P(t)$?

Attempt

If we take $P(t)$ as common factor (I'm not sure I'm doing this correctly though) ,then we have: $$ \dot{P}(t)=A^TP(t)+P(t)A=(A^T+A)P(t) $$ and using the same rationale in the background section, we have $$ P(t)=e^{(A^T+A) t}P(0)=e^{A^T t}P(0)e^{At} $$ Is this a correct solution?

Note

I know the answer is $P(t)=e^{A^T t}P(0)e^{At}$ but I'm not sure how to find it.

Answer 1

For small time increment you get approximately $$ P(t+dt)=P(t)+(A^TP(t)+P(t)A)dt+O(dt^2)=(I+A^T\,dt)P(t)(I+A\,dt)+O(dt^2) $$ which means that from time step to time step you get an accumulation of these factors on both sides, $$ P(N\,dt)=(I+A^T\,dt)^NP(t)(I+A\,dt)^N+O(Ndt^2) $$ Now if $N\,dt=\Delta t$ we get, using $(I+B/N)^N=\exp(B)+O(B^2/N)$ $$ P(t+Δt)=e^{A^T\,Δt}P(t)e^{A\,Δt}+O(Δt\,dt) $$ so that for $dt\to 0$ one gets exactly the claimed solution form.

Answer 2

By vectorizing $P$ and using the Kronecker product, similar to a method which can be used to solve Sylvester equations, then the differential equation can also be written as

$$ \frac{d}{dt}\,\text{vec}\,P(t) = \underbrace{\left(I \otimes A^\top + A^\top\otimes I\right)}_{M}\,\text{vec}\,P(t), \tag{1} $$

which has the solution

$$ \text{vec}\,P(t) = e^{M\,t}\,\text{vec}\,P(0). \tag{2} $$

By using the mixed-product property of the Kronecker product it can be shown that $I \otimes A^\top$ commutes with $A^\top \otimes I$. This commuting property allows you to write the matrix exponential as

$$ e^{M\,t} = e^{(I\,\otimes\,A^\top)\,t}\,e^{(A^\top\,\otimes\,I)\,t}. \tag{3} $$

By using the definition of a matrix exponential and again the mixed-product property of the Kronecker product then is can also be shown that

$$ e^{X\,\otimes\,I} = e^{X} \otimes I, \quad e^{I\,\otimes\,X} = I \otimes e^{X}, $$

thus

$$ e^{M\,t} = \left(I \otimes e^{A^\top t}\right) \left(e^{A^\top t} \otimes I\right). \tag{4} $$

Substituting $(4)$ into $(2)$ gives

\begin{align} \text{vec}\,P(t) &= \left(I \otimes e^{A^\top t}\right) \left(e^{A^\top t} \otimes I\right) \text{vec}\,P(0) \\ &= \left(I \otimes e^{A^\top t}\right) \text{vec}\left(P(0)\,e^{A\,t}\right) \\ &= \text{vec}\left(e^{A^\top t}\,P(0)\,e^{A\,t}\right) \end{align}

thus

$$ P(t) = e^{A^\top t}\,P(0)\,e^{A\,t}. $$

Answer 3

Consider the differential equation $$ \dot{P}(t)=A^TP(t)+P(t)A $$

We can rewrite it as: $$ \dot{P}(\tau)-A^TP(\tau)-P(\tau)A=0 $$

Multiply the last differential equation by $e^{-A^T\tau}$ from the left and by $e^{-A\tau}$ from the right then

\begin{equation} e^{-A^T\tau}\dot{P}(\tau)e^{-A\tau}-e^{-A^T\tau}A^TP(\tau)e^{-A\tau}-e^{-A^T\tau}P(\tau)Ae^{-A\tau}=0 \quad (1) \end{equation} We can easily see that equation $(1)$ is generated by the derivative of three multiplied functions: \begin{equation} \frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)=e^{-A^T\tau}\dot{P}(\tau)e^{-A\tau}-A^Te^{-A^T\tau}P(\tau)e^{-A\tau}-e^{-A^T\tau}P(\tau)Ae^{-A\tau}=0 \end{equation}

So we come up with:

$0=\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)$

Take the integral of both sides

\begin{equation} 0=\int_0^t\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)d\tau=\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)\bigg]_0^t=e^{-A^Tt}{P} \tau)e^{-At}-e^{0}{P}(0)e^{0}\end{equation} \ \begin{equation} 0=e^{-A^Tt}{P}(t)e^{-At}-P(0) \end{equation}

Multiply left and right sides by $e^{A^Tt}$ and $e^{At}$ respectively, we get:

\begin{equation} P(t)=e^{A^Tt}{P}(0)e^{At} \end{equation}

Note: I used the fact that $e^{-A^T\tau}$ and $A^T$ commute, i.e. $A^Te^{-A^T\tau}=e^{-A^T\tau}A^T$