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Homework Help: Solve du/dt=Pu when P is a projection

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve du/dt=Pu when P is a projection.

    [1/2 1/2 = du/dt with
    1/2 1/2]

    [5 = u(0).

    Part of u(0) increases exponentially while the nullspace part stays fixed.

    2. Relevant equations

    du/dt = Au with u=u(0) at t=0

    3. The attempt at a solution

    I'm stuck with this problem. I was thinking that u(t)=eλt*x.

    Would du/dt be considered matrix A??? Thanks!
  2. jcsd
  3. Oct 27, 2011 #2


    Staff: Mentor

    Re: Eigenvalues

    I don't understand much of what you have written.
    Is this P?
    [1/2 1/2]
    [1/2 1/2]

    If so, it's not a projection matrix - it is just 1/2 I, the identity matrix.

    du/dt is a two-dimensional vector, whose components are du1/dt and du2/dt.

    Your system of differential equations, not written in matrix form would look something like this:

    du1/dt = a11u1 + a12u2
    du2/dt = a21u1 + a22u2

    This could be written in vector/matrix form as du/dt = Au.
  4. Oct 27, 2011 #3
    Re: Eigenvalues

    That's where I'm confused at.

    [1/2 1/2 = du/dt. So I guess this means that the matrix equals Pu.
    1/2 1/2]'

    I'm not sure what to make of the matrix.
  5. Oct 27, 2011 #4


    Staff: Mentor

    Re: Eigenvalues

    That doesn't make any sense to me. What is the exact wording of the problem?
  6. Oct 27, 2011 #5

    I like Serena

    User Avatar
    Homework Helper

    Re: Eigenvalues

    Hi tatianaiistb! :smile:

    [1/2 1/2]
    [1/2 1/2]
    is the matrix for the orthogonal projection on (1,1).
    (Sorry Mark :redface:)

    So with u(t)=(x(t), y(t)) the problem reads:
    [tex]\begin{align}\dot x ={1 \over 2}x + {1 \over 2}y \\ \dot y ={1 \over 2}x + {1 \over 2}y\end{align}[/tex]
    [tex]\begin{align}x_0 = 5 \\ y_0 = 3\end{align}[/tex]

    Can you solve that?
  7. Oct 27, 2011 #6
    Re: Eigenvalues

    Thank you both... I think I figured out the problem. I greatly appreciate it!
  8. Oct 27, 2011 #7


    Staff: Mentor

    Re: Eigenvalues

    No problem and no need for apology. I wasn't able to make heads or tails out of what she was trying to do, so I'm glad you jumped in.
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