I am currently starting to solve numerically a system of differential equations like this:


Here, the matrices are of dimension $d\times d$. $S(A)$ is some superoperator that acts on $A(t)$. What is important in my case is that $A$ is always hermitian, which means that I do in principle only need to solve the differential equations for e.g. the elements given by the upper triangular and get the rest from hermitian property. That would allow to solve only $\frac{1}{2}(d^2+d)$ equations instead of $d^2$ which might give some noticable speedup and also reduces memory consumption. I am wondering if NDSolve can automatically take account of this or if I have to feed it with the reduced system and put everything together manually.

I searched the doc but didn't see anything like this.

Details about reduction

To clarify what I mean with a manual reduction consider $\begin{pmatrix}\dot{a}_{11} & \dot{a}_{12}\\ \dot{a}_{21} & \dot{a}_{22}\end{pmatrix} = \begin{pmatrix} a_{11}- a_{21} & 1+ a_{12}\\ a_{22}-3 a_{21} & a_{11}\end{pmatrix}$

which is completely arbitrary. But since $a_{21}= a_{12}^*$ one can just replace the $ a_{21}$ on the right handside, solve the differential equations given by the upper triangle and obtain the rest (in this example $ a_{21}$ as the complex conjugate of the solution just found.

  • $\begingroup$ I think the answer to your question is no. But I am not sure I understand you statement about doing it manually; how would you go about that? $\endgroup$ – user21 Oct 7 '15 at 5:59
  • $\begingroup$ @user21 What I mean is that since $A=(a_{ij})$ is hermitian, it is valid to only look at the upper triangular matrix of the left and right handside of the differential equation. That is, replace all $a_{ji}$ with $a_{ij}^*$, solve for them and that's it. So this is some “manipulation“ I will then probably do byby hand in advance and feed the reduced system to NDSolve $\endgroup$ – Lukas Oct 7 '15 at 6:33
  • $\begingroup$ Hm but is the aji not a∗ij to begin with - that's what makes it hermitian in the first place. But maybe I just need to get a coffee... $\endgroup$ – user21 Oct 7 '15 at 6:40
  • $\begingroup$ @user21 Isn't that what I wrote? Except that the conjugate appears after the indices ;-) $\endgroup$ – Lukas Oct 7 '15 at 11:47
  • $\begingroup$ Unfortunately, I am failing to understand that part of the question (the manual reduction) so I can not help here, but that does not seem terribly relevant as you seem to have an idea what needs to be done. If not, perhaps a more concrete example would help. $\endgroup$ – user21 Oct 7 '15 at 12:07

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