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I have the following system of differential equations: $ \frac{d}{dt} \left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c c} -(a+b) & 0\\ a & -(a+b)\\ \end{array} \right] \left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] $

I believe I have correctly obtained the eigenvalue to be $\lambda = -(a+b)$ and associated eigenvector to be $\nu = \left[ \begin{array}{c} 0\\ 1\\\end{array} \right]$.

I am trying to solve the system by the following:

$\left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c} \nu & \nu\end{array} \right] \left[ \begin{array}{c c} \exp^{\lambda t} & 0\\ 0 & \exp^{\lambda t}\\ \end{array} \right]\left[ \begin{array}{c} A(0)\\ N(0)\\\end{array} \right]$

$\left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c c} 0 & 0\\ 1 & 1\\ \end{array} \right] \left[ \begin{array}{c c} \exp^{-(a+b)t} & 0\\ 0 & \exp^{-(a+b)t}\\ \end{array} \right]\left[ \begin{array}{c} A(0)\\ N(0)\\\end{array} \right]$

$\left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = \left[ \begin{array}{c} 0\\ A(0)\exp^{-(a+b)t}+N(0)\exp^{-(a+b)t}\\\end{array} \right]$

Clearly I am doing something wrong. I suspect it has something to do with the fact that there is only one eigenvalue for this system. I was hoping Someone could help me figure this out?

Thanks

Mark

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1 Answer 1

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Hints:

  • You found the correct eigenvalue and it is a repeated eigenvalue $\lambda_{1,2} = -(a+b)$.
  • You need to find two linearly independent eigenvectors.
  • $[A-\lambda I]v = 0$ only produces one linearly independent eigenvector, so you must find a generalized one.
  • The first eigenvector you found is correct.
  • Find a linearly independent generalized second eigenvector.
  • For the second eigenvector, try $[A-\lambda I]v_2 = v_1$.

Once you have found that eigenvector, you can write the solution as (because of the repeated eigenvalue):

$$X(t) = \left[ \begin{array}{c} A(t)\\ N(t)\\\end{array} \right] = c_1~e^{\lambda~t}~(t~v_1+v_2) + c_2~e^{\lambda~t}~v_1$$

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  • $\begingroup$ Thank you. I've tried this, but it doesn't seem to work. Shouldn't the formula for the second eigenvector equate to zero? $\endgroup$
    – MPahuta
    Nov 17, 2013 at 15:09
  • $\begingroup$ No, when you do that, it is not going to produce an independent second eigenvector. Review tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx. This is what I show above, it is the process to find another eigenvector. $\endgroup$
    – Amzoti
    Nov 17, 2013 at 15:44
  • $\begingroup$ Great work! +1 (I just provided a link to the same site, different content!) $\endgroup$
    – amWhy
    Nov 17, 2013 at 16:48
  • $\begingroup$ Great, thank you, finally got it! Is there a textbook I could cite for this technique? I am a biologist and have no formal training in matrix algebra. So I need to back-up what I do. $\endgroup$
    – MPahuta
    Nov 18, 2013 at 4:03
  • $\begingroup$ A beautiful book with a slant toward economics and biology is "Dynamical Systems" by Pierre N.V. Tu. I also answered your other question, did you see it? Please make sure to review upvote and / or accept that answer also if it is helpful. Regards $\endgroup$
    – Amzoti
    Nov 18, 2013 at 4:08

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