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Math Help - Eigenvectors and Lorenz Equations

  1. #1
    Newbie LeraysJeans's Avatar
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    Question Eigenvectors and Lorenz Equations

    Hey MHF,

    So I am working through the Lorenz Equations and have run into a road block that has stumped me all day, I have turned to my beloved internet for help again.


    So I have managed to work through the system of equ.

    and have used θ=10 and b=8/3... I know original... which gives me this matrix:::


    Now I get stumped on producing the right the eigenvectors, which is a laborious process most of the time but it was beyond me today...
    anyways, Eigenvalues are
    λ= -8/3
    λ= (-11-(81+40r)^(1/2))/2 and
    λ= (-11+(81+40r)^(1/2))/2
    Duhhh... this is easy but when I use λ= -8/3
    I do not get the ε=(0,0,1) which my solutions say I should get....
    Instead I get something like:
    22/3ε1=10ε2
    I am so very lost....Could someone please point me in the right direction.
    Attached Thumbnails Attached Thumbnails Eigenvectors and Lorenz Equations-lorenz2.jpg   Eigenvectors and Lorenz Equations-lorenz1.png  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Eigenvectors and Lorenz Equations

    Quote Originally Posted by LeraysJeans View Post
    Duhhh... this is easy but when I use λ= -8/3
    I do not get the ε=(0,0,1) which my solutions say I should get....
    We have:

    E=\ker (A+(8/3)I)\equiv\begin{Bmatrix}(-22/3)x_1+10x_2=0\\ rx_1+(5/3)x_2=0\\0=0\end{matrix}

    So \epsilon=(0,0,1)^t is a non null solution of the system, as a consequence \epsilon is an eigenvector associated to \lambda=-8/3 . Notice that \{\epsilon\} is a basis of E (dimension 1) if and only if D=\det \begin{bmatrix}{-22/3}&{10}\\{r}&{5/3}\end{bmatrix}\neq 0 , if D=0 we have a basis \{\epsilon\;,\;\epsilon_1\} (dimension 2).
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  3. #3
    A Plied Mathematician
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    Re: Eigenvectors and Lorenz Equations

    Just a random, stupid thought here, but aren't the Lorenz equations nonlinear? Are you working with a linearization? Or are you trying to use a technique that is really only appropriate for linear systems?
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  4. #4
    Newbie LeraysJeans's Avatar
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    Re: Eigenvectors and Lorenz Equations

    Quote Originally Posted by FernandoRevilla View Post

    D=\det \begin{bmatrix}{-22/3}&{10}\\{r}&{5/3}\end{bmatrix}\neq 0 , if D=0 we have a basis \{\epsilon\;,\;\epsilon_1\} (dimension 2).
    FernandoRevilla! Thank you! The determinate of the matrix was the icing on the cake.

    Quote Originally Posted by Ackbeet View Post
    Just a random, stupid thought here, but aren't the Lorenz equations nonlinear? Are you working with a linearization? Or are you trying to use a technique that is really only appropriate for linear systems?
    Thanks for the concern Ackbeet, I forgot to include the fact that I input Lorenz's equations into a Jacobian Matrix. Which looks at locally linear systems around point (x,y,z). My bad. I'm always skipping steps...
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  5. #5
    A Plied Mathematician
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    Re: Eigenvectors and Lorenz Equations

    Quote Originally Posted by LeraysJeans View Post
    Thanks for the concern Ackbeet, I forgot to include the fact that I input Lorenz's equations into a Jacobian Matrix. Which looks at locally linear systems around point (x,y,z). My bad. I'm always skipping steps...
    Jolly good. Thanks for the info.
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