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Math Help - Solution by Characteristic Value

  1. #1
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    Solution by Characteristic Value

    Heres a problem that I have solved:

    Find the general solution of the equation x' = Ax, where A is the given matrix:

    [-2 4]
    [1 1]


    Using characteristic values, this matrix turns into:

    [-2-λ 4]
    [1 1-λ]

    Simplifies to (λ + 3)(λ - 2) = 0

    λ1 = 2
    λ2 = -3

    For λ1, (A - λI)k = 0
    [-4 4][k1]=[0]
    [1 -1][k2]=[0]

    Solving: c1e^(2t)[1]
    [1]

    Set this up the same way with λ2:

    [1 4][k1]=[0]
    [1 4][k2]=[0]

    Solving: c2e^(-3t)[4 ]
    [-1]


    Thus the solution is x(t) = c1c1e^(2t)[1] + c2e^(-3t)[4 ]
    [1] [-1]




    How do I go about solving for the general solution of the matrix:

    [-2 0]
    [0 -2]

    I run into a brick wall pretty quick on this one.
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  2. #2
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    Gdansk, Poland
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    In this case we have multiple eigenvalue. There is the method which deals with such cases. Generally it relays on solving equation (A-lambda*I)^n x = 0

    Maybe I will write something tomorrow about it....

    Try looking here: http://www.mif.pg.gda.pl/homepages/a...kladRownan.pdf

    I guess you don't speak polish, but equations are international.
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  3. #3
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    Hey thanks for the response. I did take a look at the "polish" document, but I don't think it addressed my pickle:

    [-2 0]
    [0 -2]


    Using characteristic values this turns into:

    [-2-λ 0]
    [0 -2-λ]

    Thus:

    (-2-λ)^2 = 0

    λ1 = -2, λ2 = -2

    So:

    [-2--2 0]
    [0 -2--2]

    [0 0][k1]=[0]
    [0 0][k2]=[0]


    So this is where I'm stuck.
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  4. #4
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    Gdansk, Poland
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    This case is very simple. We have equation:

    \dot{\mathbf{x}} = \mathbf{A}\mathbf{x}

    for
    \mathbf{A} = \left[\begin{array}{lr} -2 & 0 \\ 0 & -2 \end{array}\right]

    The case is very simple. If \mathbf{x} = [x_1, x_2]^T than the equation reduces to:
    \begin{array}{l} \dot{x_1} = -2 x_1 \\ \dot{x_2} = -2 x_2 \end{array}

    For which obvious solution is
    \begin{array}{l} x_1 = C_1 e^{-2t} \\ x_2 = C_2 e^{-2t} \end{array}

    Hence \mathbf{x} = [C_1, C_2]^T e^{-2t}
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  5. #5
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    Gdansk, Poland
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    Of course you want to use eigenvalues at all costs. Than the case complicates.

    The problem is we have multiple eigenvalue. The general formula for fundamental matrix I mentioned before is:
    (*) <br />
e^{\mathbf{A}t} = \sum_{i = 1}^{n} \sum_{j = 0}^{k_i} e^{\lambda_i t} \frac{t^j}{j!}(\mathbf{A} - \lambda_i \mathbf{I} )^{j} \mathbf{E}_{\lambda_i} <br />

    Where n is eigenvalues number, k_i is multiplicity of i-th eigenvalue and \mathbf{E}_{\lambda_i} is projection matrix from \mathbb{R}^m to

    span\{v_{\lambda_i}^{(k)}\}_{k=1}^{k_i}

    Where \{v_{\lambda_i}^{(k)}\}_{k=1}^{k_i} are generalised eigenvectors which are solution of the equation:
    (\mathbf{A} - \lambda_i \mathbf{I} )^{k_i} \mathbf{x} = 0

    In our case span\{v_{-2}^{(k)}\}_{k=1}^{2} = \mathbb{R}^2 (up to isomorphism). Hence projection matrix \mathbf{E}_{-2} = \mathbf{I}

    Using formula (*) we should assume 0^0 = 1 otherwise it does not make sense... We get (only one part in sum):

    e^{\mathbf{A}t} = \mathbf{I} e^{-2t}

    Thus:

    \mathbf{x} = [C_1, C_2]^T e^{\mathbf{A}t} =  [C_1, C_2]^T\mathbf{I} e^{-2t} = [C_1, C_2]^T e^{-2t}

    We got the previous result!
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