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Differential equations and equilibrium points

  1. Jun 7, 2009 #1
    1. The problem statement, all variables and given/known data


    http://img195.imageshack.us/img195/4873/46343978.jpg [Broken]


    3. The attempt at a solution

    Part (a) is simple enough, i get

    [tex]\dot{x}=y[/tex]
    [tex]\dot{y}=x^{3}-x[/tex].

    Equilibrium points occur when the time-derivatives of both x and y are zero, which gives the 3 equilibrium points (0,0), (1,0) and (-1,0).

    Now I thought i'd better write it as a matrix equation as this is how i remember doing these type of problems, so if we write

    [tex]X=\begin{pmatrix}x \\ y \end{pmatrix}[/tex] then

    [tex]\dot{X}=\begin{pmatrix} 0 & 1 \\ x^{2}-1 & 0 \end{pmatrix}X[/tex]

    (sorry, you have to look closely to see where the dots are!).

    I would have thought i'd then have to find the eigenvales of this matrix for each of my equilibrium points to find the stability of each one? But when I look at the solution, it uses the Jacobian matrix and finds the eigenvalues of that instead?! I'm confused!

    Thanks for any help!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 7, 2009 #2

    CompuChip

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    I don't think you can really do what you did, because x is now both in the vector X and in the coefficient matrix.

    IIRC the idea is usually that you linearize the equation around the equilibrium points. So you make a first-order Taylor series around the equilibrium point. The constant term will simply be the equilibrium point itself, and the first order term (which gives the information about the stability) will be the Jacobian multiplied by a vector. Try working it out for yourself.
     
  4. Jun 7, 2009 #3
    Ah okay thanks. I'll have a go.
     
  5. Jun 7, 2009 #4

    HallsofIvy

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    You can't write the two equations as a matrix equaiton because it is not linear. What you can do is approximate it at each point of interest (in particular, the equilibrium points) as a linear problem. You should remember that when you approximate a function, at a point, by a linear function you are using the tangent line approximation so that the slope is the derivative.

    That is what is happening here. To get a linear equation [itex]\dot{X}= Ax[/itex], A is the "derivative", that is, the Jacobian you mention, evaluated at each equilibrium point.

    By the way, the derivative of [itex]x^3- x[/itex] is [math]3x^2- 1[/math], not [math]x^2- 1[/math]!
     
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