suppose I have "n" coupled differential equation represented by the matrix,

Y = A Y

, where Y is the column matrix containing first derivatives, namely, y1(t), y2(t), ... yn(t) . A is a square matrix whose each element contains some function dependent on "t" (not constants) and Y is the column matrix containing the solution set, namely, y1(t), y2(t), ... yn(t) .

If A, contained constants, then its easy to solve by Matrix Exponential method or Eigen-Value method. But, if it contains some varying functions, then is there any approach to solve this. Please, direct me to a good reference, if possible, with an example.

  • $\begingroup$ Sounds kind of like covariant differentiation. $\endgroup$ – ben Oct 9 '12 at 17:59
  • $\begingroup$ @ben am sorry, I didn't get u, but actually this is not a derivative, but a set of coupled differential equations, represented in a matrix form. But, till now I have seen the case for which 'A' is a square matrix containing constants, but what about the case if 'A' is a square matrix, which are functions of 't' $\endgroup$ – Sai Nikhil Oct 9 '12 at 18:05

If your matrices commute at different times, that is $A(t)A(s) = A(s)A(t)$, then the solution is $Y(t) = \exp \Big( \int_0^t A(s) \mathrm{d}s \Big) Y(0)$. Here $\exp$ is just an ordinary matrix exponential.

If they do not commute, then you might be able to use Dyson formula (Dyson series). In the realm of quantum mechanics you have the following proposition:

Let $\mathscr{H}$ be a Hilbert space, $A: \mathbb{R} \to \mathscr{B}(\mathscr{H})$ a strongly continuous Hermitian operator valued function (i.e. each $A(t)$ is bounded and Hermitian). Then there is a unique solution of $$ i \partial_t \psi(t) = A(t) \psi(t), \quad \psi(0) \in \mathscr{H}. $$ If $A$ is continuous w.r.t. the operator norm, then the solution is of the form $$ \psi(t) = T \exp \Big( - i \int_0^t A(\tau) \mathrm{d}\tau \Big) \psi(0), $$ where $T \exp$ is the so called "time-ordered exponential".

I suspect that the assumption of Hermiticity is not important (you need it for unitarity of the solution). Boundedness, on the other hand, is essential. But this is no problem in yout case of a matrix.

  • $\begingroup$ Wow, its great that you've tried to help me. Million Thanks for that. Can you please provide me a reference for above discussion, so that I can look into it in more detail. And also can you please tell me what Dyson series is. I'm very new to it, but very much acquainted with Taylor series though. Thanks again for your attempt(a whole-hearted appreciation). $\endgroup$ – Sai Nikhil Oct 9 '12 at 23:20
  • $\begingroup$ excuse me sir, is it "0" in place of "s" in the above expression like Psi(t) = Texp(-i*int(A(tau).d(tau))_0-to-t)*Psi(0) ? $\endgroup$ – Sai Nikhil Oct 9 '12 at 23:48
  • $\begingroup$ I have improved the answer above. Hope it helps. I am not able to give you additional exact reference as I am away from my office. But I guess this should be a fairly standard stuff from any textbook on mathematical methods of quantum mechanics. I would also recommend you to look for some general theory of differential equations in Banach spaces. I will try to give you some references soon. $\endgroup$ – kalvotom Oct 9 '12 at 23:50
  • $\begingroup$ also, please provide me a reference, where can I get the background knowledge for it. Because, I myself, am new to Matrix Algebra, but quite good @ Calculus. $\endgroup$ – Sai Nikhil Oct 9 '12 at 23:51
  • $\begingroup$ Yes, it was a typo. You can have a different value of the inital time, though. $\endgroup$ – kalvotom Oct 9 '12 at 23:51

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