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Thread: Using eigenvectors and eigenvalues to solve DE in matrix form

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    9

    Using eigenvectors and eigenvalues to solve DE in matrix form

    Hi, my friend asked for help with a question and I'm wondering if there's a better method than mine by the question wording.

    We have the matrix
    $\displaystyle A =\left( \begin{array}{cc}
    2 & 1 \\
    1 & 2
    \end{array}\right)$
    and have to find the eigen values and vectors which are
    3 with $\displaystyle \left( \begin{array}{c}
    1 \\
    1
    \end{array} \right)$
    and
    1 with $\displaystyle \left( \begin{array}{c}
    -1 \\
    1
    \end{array}\right)$

    The next part is to solve the set of equations

    $\displaystyle u_1'' = 2 u_1' + u_2'$
    $\displaystyle u_2'' = u_1' + 2 u_2' $
    $\displaystyle u_1(0) = 0$
    $\displaystyle u_1'(0) = 1$
    $\displaystyle u_2(0) = 1$
    $\displaystyle u_2'(0) = 0$
    using the eigenvalues and eigenvectors.

    As this is of the matrix form I solve this by writing characteristic equations with the eigenvalues as the exponentials
    $\displaystyle u_1 = A e^{t} + B e^{3t} + C$
    $\displaystyle u_2 = D e^{t} + E e^{3t} + F$

    and substituting into the above equations to get six equations in A,B,C,D,E and F thus leading to
    $\displaystyle u_1 = \frac{1}{6}\left( 3e^{t} + e^{3t} -4 \right)$
    $\displaystyle
    u_2 = \frac{1}{6}\left( -3e^{t} + e^{3t} +8 \right)
    $
    which is right!

    However, I haven't used the eigenvectors only the eigenvalues, is there another (better ? ) method that I should be using?

    Thanks!
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    You can write the set of equations as
    $\displaystyle U'' = AU'$ where $\displaystyle U = \left( \begin{array}{c}
    u_1 \\
    u_2
    \end{array} \right)$

    A is diagonalisable into the diagonal matrix $\displaystyle D =\left( \begin{array}{cc}
    3 & 0 \\
    0 & 1
    \end{array}\right)$ through a transition matrix P such that $\displaystyle P^{-1}AP = D$

    Let $\displaystyle V = \left( \begin{array}{c}
    v_1 \\
    v_2
    \end{array} \right) = P^{-1}U \Rightarrow U = PV$

    $\displaystyle U'' = AU'$ is transposed into $\displaystyle PV'' = APV' \Rightarrow V'' = P^{-1}APV' = DV'$

    Since D is diagonal $\displaystyle V'' = DV'$ is easily solved.

    Then $\displaystyle U = PV$ is used to find $\displaystyle u_1$ and $\displaystyle u_2$
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