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Problem in a system of ODEs

  1. Jun 15, 2012 #1
    I'm running into a problem. This is mainly for reading over the summer and I'm working on getting through a dynamical systems book on my own. I've come across a system that I'm not too sure on the procedure.

    Consider the following system of differential equations:
    [itex]\frac{dX}{dt} = 1 - X - XY - XZ, \\
    \frac{dY}{dt} = aXY - Y,\\
    \frac{dZ}{dt} = aXZ - Z.
    [/itex]

    Here, 'a' is a real positive constant and [itex] X,Y,Z \geq 0 . [/itex]

    I can find all of the three steady states except for the one where all three exist, i.e. when [itex] \bar{X},\bar{Y},\bar{Z} \neq 0 . [/itex]

    When solving for this steady state, I arrive at [itex] \bar{X} = \frac{1}{a} . [/itex]

    Then, plugging this back into the first equation, I get the following

    [itex] 0 = a -1 - \bar{Y} - \bar{Z}, [/itex]

    or,

    [itex] a-1 = \bar{Y}+\bar{Z}. [/itex]

    Is there a way to solve for [itex]\bar{Y}[/itex] and [itex]\bar{Z}[/itex]?

    -as a side note, I've explored this numerically with matlab and it looks like it depends on the initial conditions. Could it be a saddle?
     
  2. jcsd
  3. Jun 15, 2012 #2
    Continuing to the Jacobian anyways, we arrive at

    [itex]
    J = \left[
    \begin{array}{ccc}
    -1 - (\bar{Y} + \bar{Z}) & -\bar{X} & -\bar{X} \\
    a\bar{Y} & a\bar{X}-1 & 0 \\
    a\bar{Z} & 0 & a\bar{X}-1
    \end{array}
    \right].
    [/itex]

    Plugging in what we know,

    [itex]
    J = \left[
    \begin{array}{ccc}
    -a & -\frac{1}{a} &-\frac{1}{a} \\
    a\bar{Y} & 0 & 0 \\
    a\bar{Z} & 0 & 0
    \end{array}
    \right].
    [/itex]

    This means that the trace of the Jacobian is negative, i.e. [itex] tr(J) = -a < 0 [/itex]. Also that the determinant is equal to zero. My book does not talk about what happens when the determinant of the Jacobian is zero. Any insights?
     
  4. Jun 15, 2012 #3

    AlephZero

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    If you let U = Y+Z and W - Y-Z, the equations become
    dX/dt = 1 - X - XU
    dU/dt = aXU - U
    dW/dt = aXW - W

    Then, you have to consider more than one possibiltiy. From the third equation, if dW/dt = 0 then either aX = 1 or W = 0 .....
     
  5. Jun 15, 2012 #4
    Thanks, that helps me understand it a bit better.

    So, continuing that thought, if [itex] \bar{X} = \frac{1}{a} [/itex], then that implies [itex] \bar{U} = a -1,\textrm{ and } \bar{W} = 0. [/itex] This means that [itex] \bar{Y} = \bar{Z}. [/itex]

    So [itex] a-1 = 2 \bar{Y}. [/itex] Hence [itex] \bar{Y} = \bar{Z} = \frac{a-1}{2} [/itex].

    But I'm still confused about the stability. The Jacobian of the new system is technically only slightly different but still results in a determinant of zero and a trace of [itex] -a [/itex].

    Looking at the characteristic equation of such a matrix, we get

    [itex] det[J-I\lambda] = -\lambda^{3} + \lambda^{2}\textrm{trace}[J] + \frac{1}{2} \lambda \left[ \textrm{trace}[J^{2}] - \textrm{trace}[J]^{2} \right] + \textrm{det}[J]. [/itex]

    The trace and determinant are the same for both systems. I just hand wrote out the square of the Jacobian and the trace of it comes out to be just [itex] a^{2} [/itex].

    So the characteristic equation is

    [itex] det[J-I\lambda] = -\lambda^{3} - a \lambda^{2} = 0 [/itex],

    and

    [itex] \lambda_{1,2}= 0, \textrm{ and } \lambda_{3} = -a [/itex].

    And all the rules I know only apply when the eigenvalues are non-zero. I've read what two books I have on this and they don't go into this situation.

    Any thoughts on the stability?
     
  6. Jun 23, 2012 #5

    pasmith

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    Homework Helper

    I think you've fallen into error here.

    For a fixed point, all of [itex]\dot X[/itex], [itex]\dot Y[/itex] and [itex]\dot Z[/itex] must vanish. Looking at the Y equation, we see that [itex]Y = 0[/itex] or [itex]aX-1 = 0[/itex]. Looking at the Z equation, we see that [itex]Z = 0[/itex] or [itex]aX-1 = 0[/itex].

    Taking the case [itex]Y = Z = 0[/itex], the X equation gives [itex]X = 1[/itex], so there is a fixed point at [itex](1,0,0)[/itex].

    In the case [itex]aX-1=0[/itex], there will be a fixed point if [itex]Y + Z = a - 1[/itex]. Thus every point on the line [itex](X,Y,Z) = (\frac{1}{a}, s, a-1-s)[/itex] is a fixed point.

    (Making the substitutions proposed by AlephZero gives [itex]U = Y + Z= a -1[/itex] and [itex]X = \frac{1}{a}[/itex], but [itex]W = Y - Z[/itex] is arbitrary, not 0 as you have assumed.)

    The characteristic polynomial of the Jacobian is then
    [tex]0=\mathrm{det}(J - \lambda I) =
    \left|\begin{array}{ccc}-a-\lambda & -a^{-1} & -a^{-1} \\
    aY & -\lambda & 0 \\
    aZ & 0 & -\lambda \end{array}\right| =
    (-a-\lambda)\left|\begin{array}{cc} -\lambda & 0 \\ 0 & -\lambda \end{array} \right|
    +a^{-1} \left|\begin{array}{cc} aY & 0 \\ aZ & -\lambda \end{array} \right|
    -a^{-1} \left|\begin{array}{cc} aY & -\lambda \\ aZ & 0 \end{array}\right|
    [/tex]
    which reduces to
    [tex]
    \lambda^2(\lambda+a)+\lambda Y + \lambda Z =
    \lambda(\lambda^2 + a\lambda + (a-1)) =
    \lambda(\lambda + 1)(\lambda+ a - 1) = 0
    [/tex]
    There is therefore one zero eigenvalue (corresponding to the line of fixed points).

    In general, when the jacobian at a fixed point has a zero eigenvalue, there are two possibilities. The first is that the system is degenerate and has non-isolated fixed points. The second is that the fixed point is not hyperbolic, and you must look at higher-order terms to determine the stability of the fixed point (the key term is "center manifold").

    Here we have the first case: the system is degenerate. Stability is therefore determined by the other two eigenvalues: -1 (stable) and [itex]1 - a[/itex] (unstable if [itex] a < 1[/itex], indeterminate if [itex]a = 1[/itex], and stable if [itex]a > 1[/itex]).
     
  7. Jun 23, 2012 #6

    pasmith

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    Now I'm up to 10 posts, I can give you an actual link: Center manifold
     
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