Solving Logistic ODE with Non-commuting Matrices

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SUMMARY

The discussion focuses on finding a general solution for the logistic ordinary differential equation (ODE) represented as \(\frac{dU}{dx}=A(I-U)U\), where \(A\) and \(U\) are square matrices. The proposed solution \(U=(I+e^{-Ax})^{-1}\) is valid only when \(U\) and \(A\) commute. The general scalar solution is given by \(\frac{E^{Ax}}{E^{Ax} + E^C}\), but complications arise when \(U\) is a function of both \(A\) and the initial condition, leading to non-commutation issues. The discussion also explores diagonalization of \(A\) and the initial condition as a potential approach to simplify the problem.

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Manchot
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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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