I'm trying to find a general solution for the logistic ODE [itex]\frac{dU}{dx}=A(I-U)U[/itex], where A and U are square matrices and x is a scalar parameter. Inspired by the scalar equivalent I guessed that [itex]U=(I+e^{-Ax})^{-1}[/itex] is a valid solution; however, [itex]U=(I+e^{-Ax+B})^{-1}[/itex] is not when U and A don't commute. Any ideas?
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?
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^{-1}DM where D is a diagonal matrix V = MUM^{-1} The ODE becomes: dV/dx = D(I-V)V