I know how to solve the differential equation $\dot{x} = A x$ when $A$ is a constant $n\times n$ matrix. However, I cannot solve the problem when $A$ also depends on $t$.

To be more specific, $A (t)=\left(\begin{array}{cc} 1 & -1/t\\ 1+t & -1 \end{array}\right)$, where $t>0$.

I can verify that $x(t)=\left(\begin{array}{c} 1\\ t \end{array}\right)$ is a solution. However, in order to find the fundamental solution, I need another linearly independent solution. I tried to set the other solution to be $\left(\begin{array}{c} y_{1}(t)\\ y_{2}(t) \end{array}\right)$ and plugged it into the equation. Then I got the following:

$y_{1}t-y_{2}=\dot{y}_{1}t$ and $y_{1}+y_{1}t-y_{2}=\dot{y}_{2}$. Then, I was stuck.

I also tried $(\det\Phi)^{\prime}=trA\det\Phi$ and got nowhere.

  • $\begingroup$ I suppose you could try the method of Laplace transforms (I'm not being snarky, I have no idea how hard it would be here to pursue it) $\endgroup$ – James S. Cook Sep 10 '12 at 0:05

You have an ODE of the form

$$\dot{x} (t) = A (t) \, x(t)$$

which a control theorist would call a "linear time-varying (LTV) unforced system" (unforced because there is no control input). The general solution of this ODE is the following

$$x (t) = \Phi (t,t_0)\, x(t_0)$$

where the transition matrix $\Phi (t,t_0)$ is given by the Peano-Baker series (take a look at lecture 5 on Hespanha's book for more details). Since the Peano-Baker series is scary, an alternative would be to differentiate both sides of $x (t) = \Phi (t,t_0)\, x(t_0)$ with respect to time to obtain $\dot{x} (t) = \dot{\Phi} (t,t_0)\, x(t_0)$. Since we have $\dot{x} (t) = A (t) \, x(t) = A \, \Phi (t,t_0)\, x(t_0)$, we finally obtain the matrix differential equation

$$\dot{\Phi} (t,t_0) = A (t) \, \Phi (t,t_0)$$

Let $\phi_1 (t)$ and $\phi_2 (t)$ be the 1st and 2nd columns of $\Phi (t,t_0)$, respectively. Since $\Phi (t_0,t_0) = I_{2 \times 2}$, we have the initial conditions for the ODEs on $\phi_1 (t)$ and $\phi_2 (t)$.

  • $\begingroup$ Nice answer. I tried to look at the link, I guess we have to buy it? The preview I saw didn't include the relevant sections. $\endgroup$ – James S. Cook Sep 10 '12 at 0:43
  • $\begingroup$ @ James S. Cook: I can see the first page of the lecture, which is the one that contains the Peano-Baker series. $\endgroup$ – Rod Carvalho Sep 10 '12 at 0:47
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    $\begingroup$ @ James S. Cook: Here's a review of the Peano-Baker series: arxiv.org/abs/1011.1775 $\endgroup$ – Rod Carvalho Sep 10 '12 at 18:35
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    $\begingroup$ It's almost as if you tried to present this with the most jargon and symbols possible while hiding your meaning from anyone who hasn't read Rugh. Literally, you could just say the two independent solutions are 1 with an initial condition x = (1,0) and another with the initial condition x = (0,1) solving dot(x) = A(t) x. That's the same information as you posted. And seeing as he can't even come close to solving the differential equation (he said he tried to plug in y = (y_1, y_2)), that information itself is a bit missing the mark. Your advice amounts to "just solve it". $\endgroup$ – user974006 May 24 '14 at 2:58

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