# Matrix ODE

1. Jan 15, 2014

### Manchot

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?

2. Jan 15, 2014

### maajdl

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?

3. Jan 15, 2014

### maajdl

If U is a function of A,
then U commutes with A.

4. Jan 15, 2014

### Manchot

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.

5. Jan 16, 2014

### maajdl

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

6. Jan 16, 2014

### Manchot

Yeah, I tried diagonalizing both A and the initial condition. No dice.

7. Jan 16, 2014