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Diff equation by diagonalization

  1. Apr 2, 2007 #1
    1. The problem statement, all variables and given/known data
    solve initial value problem for the equation dx/dt = Ax where A =
    [1 -1]
    [0 1]
    x(0) = [1, 1]T

    x(t) = S*elambda*t*S -1*x(0)
    where S is diagonal matrix, lambda is eigenvalue;

    3. The attempt at a solution
    I tried to diagonalize it, but I get one eigenvalue =1 mult 2 and I don't have enough eigenvectors to get S!
    Am I missing something? and how do i solve it?

    Thanks
     
  2. jcsd
  3. Apr 2, 2007 #2

    D H

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    The solution to [itex]dx/dt = Ax[/itex] for scalar x is [itex]x=\exp(At)x_0[/itex]. This extends to multiple dimensions with [itex]exp(At) = \sum_r A^r t^r/r![/itex]
     
  4. Apr 2, 2007 #3
    ok, so you are saying that A does not have to be diagonalizable for a solution to exist?

    I am trying to see relations to the diagonalization
     
  5. Apr 3, 2007 #4

    HallsofIvy

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    ?? Every linear equation has solutions! Where did you get the idea that the matrix had to be diagonalizable?

    Diagonalizable makes it easier but when a matrix is not diagonalizable you can still get the "Jordan Normal form".

    The simplest way to solve this equation is two treat it as two equations:
    dx/dt= x- y, dy/dt= y, since they are "partially disconnected": you can solve for y immediately: from dy/dt= y, y= Cet. Then the second equation becomes dx/dt= x- Cet or dx/dt- x= -Cet. That has integrating factor e-t: multiplying the equation by e-t give e-tdx/dt- e-tx= d(e-tx)/dt= e-t(Cet)= C. Integrating, e-tx= Ct+ D so x= Ctet+ Det. With x(0)= D= 1 and y(0)= C= 1, we have x(t)= tet+ et, y(t)= et.

    Putting that back into matrix form that is x(t)= [tet+ et, et].
     
  6. Apr 3, 2007 #5

    D H

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    This is a linear equation of the form [itex]d\vec x/dt = \mathbf A \vec x[/itex]. If A and x were scalars, the solution would obviously be [itex]x=\exp(At)x(0)[/itex]. This is true for matrices as well:

    [tex]\vec x \exp(\mathbf At) \vec x(0)[/tex]

    where [itex]\exp(At)[/itex] is the matrix exponential

    [tex]\exp(At) = \sum_{n=0}^\infty \frac{A^n t^n}{n!}[/tex]

    Writing

    [tex]\mathbf A = \mathbf 1_{2x2} - \mathbf B[/itex]

    where

    [tex]\mathbf 1_{2x2}[/tex] is the [itex]2x2[/itex] identity matrix and

    [tex]\mathbf B \equiv \bmatrix 0 & 1 \\ 0 & 0 \endbmatrix[/tex]

    Then

    [tex]\mathbf A^2 = \mathbf 1_{2x2} - 2\mathbf B[/tex]

    and in general

    [tex]\mathbf A^n = \mathbf 1_{2x2} - n\mathbf B[/tex]

    Thus

    [tex]\exp(At) =\mathbf 1_{2x2}\sum_{n=0}^\infty \frac{t^n}{n!} - \mathbf B \sum_{n=1}^\infty \frac{t^n}{(n-1)!}[/tex]

    Simplifying,

    [tex]\exp(At) =\mathbf 1_{2x2}\exp t - \mathbf B t\exp t[/tex]

    The desired solution is

    [tex]\vec x(t) =(\mathbf 1_{2x2}\exp t - \mathbf B t\exp t)\bmatrix 1 \\ 1\endbmatrix = \bmatrix 1-t\\1\endbmatrix \exp t[/tex]
     
  7. Apr 3, 2007 #6

    D H

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    There is a sign error here. That should be d(e-tx)/dt = -C

    With the correct sign, integrating yields e-tx= D-Ct, so x= (D-Ct)et. With x(0)=y(0)=1, C=D=1, so x(t)= et-tet, y(t)= et. In matrix form,
    [tex]\vec x(t) = \bmatrix e^t-te^t \\ e^t\endbmatrix[/tex]
     
  8. Apr 3, 2007 #7

    HallsofIvy

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    Thanks for the correction. One of these days, I really need to learn arithmetic!
     
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