Solving Logistic ODE with Non-commuting Matrices

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The discussion focuses on finding a general solution for the logistic ODE involving non-commuting matrices, specifically \(\frac{dU}{dx}=A(I-U)U\). A proposed solution, \(U=(I+e^{-Ax})^{-1}\), fails when \(U\) and \(A\) do not commute. The conversation explores the implications of \(U\) being a function of both \(A\) and the initial condition, which complicates commutation. Attempts to diagonalize \(A\) and the initial condition have not yielded results. The dialogue raises questions about the necessity of a formal solution and the practical goals behind solving the equation.
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I'm trying to find a general solution for the logistic ODE \frac{dU}{dx}=A(I-U)U, where A and U are square matrices and x is a scalar parameter. Inspired by the scalar equivalent I guessed that U=(I+e^{-Ax})^{-1} is a valid solution; however, U=(I+e^{-Ax+B})^{-1} is not when U and A don't commute. Any ideas?
 
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The general solution to the scaler equation is:

E^(A x)/(E^(A x) + E^C)

where C is a constant.
Maybe this can lead to a similar solution for the matricial version?
 
If U is a function of A,
then U commutes with A.
 
I tried all sorts of versions of the scalar equation, maajdl. They all run into the same commutation problem. Unfortunately, U is a function of both A and the initial condition, which means that it doesn't commute with A unless the initial condition does.
 
Could that help?

Assuming:

A = M-1DM where D is a diagonal matrix
V = MUM-1

The ODE becomes:

dV/dx = D(I-V)V
 
Yeah, I tried diagonalizing both A and the initial condition. No dice.
 
What is your practical goal?
Why do you need a formal solution?
 
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