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Matrix ODE

by Manchot
Tags: matrix
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Manchot
#1
Jan15-14, 01:03 PM
P: 728
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?
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maajdl
#2
Jan15-14, 02:33 PM
PF Gold
P: 377
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?
maajdl
#3
Jan15-14, 02:46 PM
PF Gold
P: 377
If U is a function of A,
then U commutes with A.

Manchot
#4
Jan15-14, 06:21 PM
P: 728
Matrix ODE

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.
maajdl
#5
Jan16-14, 01:09 AM
PF Gold
P: 377
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
Manchot
#6
Jan16-14, 06:40 AM
P: 728
Yeah, I tried diagonalizing both A and the initial condition. No dice.
maajdl
#7
Jan16-14, 07:48 AM
PF Gold
P: 377
What is your practical goal?
Why do you need a formal solution?


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