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Re(eigenvalue) inequality problem

  1. Sep 28, 2005 #1


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    if a diff.eqn. has the characteristic equation [itex]\lambda^2 + (3-K) \lambda + 1 = 0[/itex]
    the eigenvalues solves to [itex]\lambda=-3/2 + K/2 \pm 1/2 \sqrt{5-6K+K^2}[/itex]. No problem there. But when is the diff.eqn. asymp. stable, meaning [itex]\Re(\lambda)<0[/itex] ?

    I can only get this far
    [itex]\Re(-3/2 + K/2 \pm 1/2*\sqrt{5-6K+K^2})<0[/itex]
    [itex]-3/2+1/2 \Re(K \pm \sqrt{5-6K+K^2})<0[/itex]

    How can i find the values for K, where this inequality is true?

  2. jcsd
  3. Sep 29, 2005 #2


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    My inclination would be to first consider the case when [tex]5-6K+K^2[/tex] is positive, and then consider the case when it is negative. If you separate them out, it shouldn't be too hard.

  4. Sep 30, 2005 #3


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    Solving K2- 6K+ 5> 0 tells us that there will be complex roots for K between 1 and 5 and real roots for K<= 1, >= 5.

    I there are complex roots, the real part is just -3/2+ K/2. That is 0 for K= 3, negative for K< 3, positive for K> 3. The solution will be stable for 1< K< 3, unstable for 3< K< 5.

    For K<= 1 or K>= 5, we need to look at all of [itex]-3/2 + K/2 \pm 1/2 \sqrt{5-6K+K^2}[/itex]

    The best way to determine where that is positive or negative is to set it equal to 0 and solve the resulting quadratic equation. Those will separate "< 0" from "> 0". Choose one value of K in each resulting interval to see whether this is positive or negative.
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