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I have two matries given by (M' means derivative w.r.t x)

$ M=\left( \begin{array}{ccc} f_1(x) & f_2(x) & f_3(x) \\ f_4(x) & f_5(x)& f_6(x) \\ f_7(x) & f_8(x) & f_9(x) \\ \end{array} \right) \tag 1$

$ M'=\left( \begin{array}{ccc} f_1'(x) & f_2'(x) & f_3'(x) \\ f_4'(x) & f_5'(x)& f_6'(x) \\ f_7'(x) & f_8'(x) & f_9'(x) \\ \end{array} \right) \tag 2$ . These $f_i(x)$ can be any type of functions. They can be equal ($f_i=f_j$)or different or equal some of them and different others ,they can be constant functions. There is no limit or condition at all. Only one exception is that total matrix M cant be a zero matrix with full entries as zeros.

Question

Can you provide all $f_i$( any function you can select) such that it just satisfies following condition(all $a_0,b_0,c_0,a_1,b_1,c_1$ are constants cant be altered,but x is a variable) . or is it not possible to have such one M? please clarify

$ M'=\{\left( \begin{array}{ccc} 0 & -c_0 & b_0 \\ c_0 & 0 & -a_0 \\ -b_0 & a_0 & 0 \\ \end{array} \right)+\left( \begin{array}{ccc} 0 & -(c_1-c_0) & (b_1-b_0) \\ (c_1-c_0) & 0 & -(a_1-a_0) \\ -(b_1-b_0) & (a_1-a_0) & 0 \\ \end{array} \right)x\}M \tag 3$

a,b,c are constants. You can select any arbitrar function. I tried a lot. Since there is no standard methods. I couldnt get a proper one.Thanks a lot

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Not a solution but maybe a step in the right direction. First you can formulate your problem as a matrix differential equation. With the matrix vectorization operator, which stacks the columns of the matrix , we write:

$$m(x) = \text{vec}(M(x))$$

whereas $m(x)$ is just the vector of your unknown functions $f_i(x)$. Your original problem is

$$M'(x) = (A + Bx)M(x)$$

which transforms to the matrix differential equation, after applying matrix vectorization on both sides

$$m'(x) = ((I \otimes A) + (I \otimes Bx))m(x)$$

with the Kronecker matrix product $\otimes$. So we have an matrix differential equation of the form

$$m'(x) = (\widetilde A + \widetilde Bx)m(x)$$

with some constant matrices $\widetilde A$ and $\widetilde B$ that essentially just contain the numbers $a_0, b_0$ etc.

This is solvable, when the matrices $\widetilde A$ and $\widetilde B$ are commuting, i.e. $\widetilde A \widetilde B = \widetilde B \widetilde A$ using the matrix exponential:

$$m(x) = \exp (\widetilde A x + 1/2\widetilde Bx^2)$$

However, I dont know whether the matrices commute (which is simple to check, but tedious) and unfortunately I have no solution when the do not commute.

This Math24 page gives some background information, on how to solve matrix differential equations with variable coefficients (as in this case).

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  • $\begingroup$ The issue is how do you evaluate $m(x) = \exp (\widetilde A x + 1/2\widetilde Bx^2)$ to check whether it satisfies the condition. Issue is when you put these exponential power especially with $Bx^2$ it is complicated $\endgroup$ – Nirvana Jul 31 '14 at 16:38
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    $\begingroup$ I slightly changed the formatting, using display math syntax $$ $$ and the link syntax [text](url). Hope you like it. $\endgroup$ – user147263 Jul 31 '14 at 16:43
  • $\begingroup$ To check the commuting condition you don't need to evaluate the matrix exponential. But I agree evaluating the matrix exponential of a 9x9 matrix is complicated. Personally I would only try it with a computer algebra package. But in case the commuting condition is met, then the solution maybe just be complicated because your problem is. Of course it also might be, that the matrix exponential simplifies significantly from the special structure of the $\widetilde{A}$ and $\widetilde{B}$ matrices. That I do not know. $\endgroup$ – Andreas H. Jul 31 '14 at 16:44
  • $\begingroup$ @Rejo_Slash See the above comment ^^^ (since I inserted mine at the same time, I'm not sure you were notified). $\endgroup$ – user147263 Jul 31 '14 at 16:45
  • $\begingroup$ Just checked now thanks $\endgroup$ – Nirvana Jul 31 '14 at 16:57

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