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Matrix ODE

  1. Jan 15, 2014 #1
    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?
     
  2. jcsd
  3. Jan 15, 2014 #2

    maajdl

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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?
     
  4. Jan 15, 2014 #3

    maajdl

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    If U is a function of A,
    then U commutes with A.
     
  5. Jan 15, 2014 #4
    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.
     
  6. Jan 16, 2014 #5

    maajdl

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    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
     
  7. Jan 16, 2014 #6
    Yeah, I tried diagonalizing both A and the initial condition. No dice.
     
  8. Jan 16, 2014 #7

    maajdl

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    What is your practical goal?
    Why do you need a formal solution?
     
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