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Consider the following system of ODEs $$ Y^{'}(t) = - \left[ A Y(t) + Y(t) A \right] + B(t) , $$ where $Y(t)$,$A$,$B(t)$ are all matrices, with the properties $A=A^T$, $Y=Y^T$. $Y(t)$ is the matrix of functions to be solved. Of course, there's a standard way to solve this system, for example, by rearranging the $n \times n$ matrix $Y(t)$ into a column vector of dimension $n^2$ and finding the eigenvalues of the resulting constant $n^2 \times n^2$ matrix. But my question is, are there any known results concerning this system such that the problem may be reduced from dimension $n^2 \times n^2$ to, say, $n \times n$? Any comment would be appreciated.

Some random observations

  1. $A Y(t) + Y(t) A$ looks like an anti-commutator. Is it related to the following fact: "the time-evolution of a quantum variable is governed by its commutator with the Hamiltonian"

  2. Is the above system related to the exponential map in Lie Algebras/ Lie Groups?

P.S. Why do I want to consider it as a generic matrix system? It's hard to answer this question. Indeed I just want to express the solution $Y(t)$ in a form that is "natural". For example, in 1-D (physical situation), the matrix $A$ is constant and band-diagonal, and equals $I + c R$, where R is the discretisation of the linear operator $\partial_x^2$. In 2-D (physical situation), $R$ is the discretisation of $\partial_x^2 + \partial_y^2$.

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    $\begingroup$ Your question is not clear enough concerning what you actually want. Also, you fail to motivate why one shouldn't view this a $n^2\times n^2$ differential system. You obtain "a lot" of properties this way. But what "lot" interests you? Also, (2) is clear enough from the previous viewpoint, so the question is: what shall the dimension of your Lie algebras/groups be? $\endgroup$
    – Loïc Teyssier
    Commented Jun 24, 2014 at 11:31
  • $\begingroup$ Thanks for replying. I've posted my response to first question as a postscript to the original question. Also you said, "(2) is clear enough from the previous viewpoint", why is that? In fact I was not trained in Lie Algebras/Lie Groups, and the observation just occurred to me as something arising from rote memory. Would you please explain that? Indeed I'm interested in obtaining a "nice" solution in closed-form for the above system, as I'm more a physicist than mathematician. $\endgroup$
    – Jamie
    Commented Jun 24, 2014 at 12:16
  • $\begingroup$ What I meant is that, up to increase dimension, the solutions will always be obtained as the Lie exponential of the (linear) vector field defining the augmented system. $\endgroup$
    – Loïc Teyssier
    Commented Jun 24, 2014 at 13:15
  • $\begingroup$ I think you're reading way too much into this. You're basically asking if every solution $Y$ of your ODE satisfies $Z:=Y-Y^t\equiv 0$. Since $Z$ solves $Z'=-ZA-AZ+B-B^t$, this (trivially) holds precisely if $B=B^t$. $\endgroup$
    – Christian Remling
    Commented Jun 24, 2014 at 19:18

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Well, there won't be solutions unless $B(t)^T = B(t)$ as well. If you have this, then consider $U(t) = e^{At}Y(t)e^{At}$. This will satisfy $$ U'(t) = e^{At}B(t)e^{At}, $$ so your solution with given initial value $Y(0)$ is going to be $$ Y(t) = e^{-At}\left(Y(0) + \int_0^te^{A\tau}B(\tau)e^{A\tau}\ \mathrm{d}\tau\right)e^{-At}. $$

By the way, as for 'reducing the dimension of the system' (which one might want to do so as to reduce the complexity of computing the matrix exponential) by 'uncoupling the system', this can be done: Writing $A = RDR^T$ where $R$ is orthogonal and $D$ is diagonal (which can always be done) and setting $Y(t) = RZ(t)R^T$ and $B(t) = RC(t)R^T$, the equation becomes $$ Z'(t) = -(DZ(t)+Z(t)D) + C(t), $$ and one recognizes this as $\tfrac12n(n{+}1)$ independent first order ODE $$ Z_{ij}'(t) = -(D_i+D_j) Z_{ij}(t) + C_{ij}(t)\qquad i\le j. $$

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  • $\begingroup$ I have to admit your method is quite clever. What you gave is exactly what I wanted. If $Y$ is considered as a quantum observable, then $e^{A t} Y e^{A t}$ is just its evolved form under the "Hamiltonian" $A$ for time $t$...of course one has to go to imaginary time, and the usual commutator becomes an anticommutator...and that explains its relation to Lie Groups. $\endgroup$
    – Jamie
    Commented Jun 24, 2014 at 14:44

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