1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\tau$.

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

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.