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

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.

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

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

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

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

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

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

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

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

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.

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

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