I have a state transition matrix $\Phi(t,\tau) = \left(\begin{matrix} e^{-(t-\tau)} & t-\tau \\ 0 & 1+t(t-\tau) \end{matrix}\right)$. I am tasked to find the state matrix $A(t)$ that corresponds to it. Here are some properties of interest about $\Phi(t,\tau)$:

  1. $\Phi^{-1}(t,\tau) = \Phi(\tau,t)$
  2. $\Phi(t,t) = \mathbf{I}$, where $\mathbf{I}$ is the identity matrix
  3. $\Phi(t,t_0)$ solves the equation $\frac{\partial}{\partial t}\Phi(t,t_0) = A(t)\Phi(t,t_0)$ with initial condition $\Phi(t_0,t_0)=\mathbf{I}$
  4. When $A(t)$ is constant (i.e., not a function of $t$), $\Phi(t,\tau)$ reduces to $e^{A(t-\tau)}$, the matrix exponential. Note that for this particular problem, $A$ may or may not be a function of $t$.

Naively, I tried to use the differential equation in property (3) in combination with property (1) to solve for $A(t)$:

$A(t) = \frac{\partial}{\partial t}\left[\Phi(t,\tau)\right]\Phi(\tau,t)$

Differentiating each term of $\Phi$ gives me $\dot{\Phi}(t,\tau) = \left(\begin{matrix} -e^{-(t-\tau)} & 1 \\ 0 & 2t-\tau \end{matrix}\right)$.

Performing the operation $\frac{\partial}{\partial t}\left[\Phi(t,\tau)\right]\Phi(\tau,t)$ gives me a function of both $t$ and $\tau$, but $A$ must be a function of $t$ only.

Any ideas as to what to do? I did not think this would be that difficult, but apparently I'm wrong.


I think your idea is correct. The problem is whether $\Phi(t,t_0)$ you give really is a state transition matrix. It doesn't satisfy the properties. For example, $\Phi(t,t)\ne I$. Even if we change $e^{-t-\tau}$ to $e^{t-\tau}$, then $\Phi(t,t) = I$, but $\Phi(t,\tau)^{-1}\ne \Phi(\tau,t)$. If $\Phi(\tau,t)$ really is a state transition matrix, $\dot{\Phi}(t,\tau)\Phi^{-1}(t,\tau)$ should cancel the term $\tau$.

  • $\begingroup$ doh :) i only checked that $\Phi(t,t) = I$, did not check the other property.. $\endgroup$ – Dang Khoa Oct 23 '11 at 2:39

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