5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Counter question: what is A? $\endgroup$ – Yuriy S Oct 16 '18 at 16:28
  • $\begingroup$ @YuriyS A is a matrix. $\endgroup$ – Lod Oct 16 '18 at 16:29
  • 2
    $\begingroup$ Dimensions are very likely mismatched in the expression $\dot{p}(t)=A^Tp(t)+p(t)A$. $\endgroup$ – Ennar Oct 16 '18 at 16:30
  • $\begingroup$ What is $p A$ then? $\endgroup$ – Yuriy S Oct 16 '18 at 16:30
  • $\begingroup$ @Ennar I fixed the question. $\endgroup$ – Lod Oct 16 '18 at 16:36
2
$\begingroup$

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. $A^Te^{-A^T\tau}=e^{-A^T\tau}A^T$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.