1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Classify the fixed points of this dynamical system

  1. Nov 29, 2015 #1
    1. The problem statement, all variables and given/known data

    $$\dot{x_1}=x_2-x_2^3,~~~~~~\dot{x_2}=-x_1-3x_2^2+x_1^2x_2+x_2$$

    I need help in determining the type and stability of the fixed points in this system.

    2. Relevant equations

    The Jordan Normal Form

    Let A be a 2x2 matrix, then there exists a real and non singular matrix M such that ##J=M^{-1}AM##, J is said to be in the jordan normal form.

    Linearization Theorem
    In the neighbourhood of a fixed point which has a simple linearization, the phase portraits of the non linear system and its linearization are qualitatively the same. This applies only if the Jacobian matrix is non-singular and have non-zero real part at the fixed point.

    3. The attempt at a solution


    The following points are fixed: ##(0,0),(-1,1) \text{ and } (2,1)##

    I will call the Jacobian matrix A:

    $$A(x_1,x_2)=\begin{bmatrix}0&&1-3x_2^2\\2x_1x_2-1&&x_1^2+1-6x_2\end{bmatrix}$$

    I will only be talking about the fixed point ##(0,0)##

    $$A(0,0)=\begin{bmatrix}0&&1\\-1&&1\end{bmatrix}$$

    The transformation matrix M consists of the eigenvectors of A: $$M=\begin{bmatrix}1&&1\\\frac{1+\sqrt{3}i}{2}&&\frac{1-\sqrt{3}i}{2}\end{bmatrix}\text{ thus, }J=\begin{bmatrix}\frac{1+\sqrt{3}i}{2}&&0\\0&&\frac{1-\sqrt{3}i}{2}\end{bmatrix}$$

    My first problem is that the I have not seen this jordan normal form before. The closest form to this that we have discussed is ##J=\begin{bmatrix}\alpha&&-\beta\\\beta&&\alpha\end{bmatrix}##, where ##\alpha## and ##\beta## are real, which is a stable focus. My second problem is that the transformation matrix M is not real, and according to the first theorem M should be real. And even if the theorem I have is incorrect, and M can be complex, Im not sure how I can transform the phase portrait back to the original coordinates using M, since it is a complex transformation. And help would be greatly appreciated! Cheers.
     
    Last edited: Nov 29, 2015
  2. jcsd
  3. Nov 30, 2015 #2

    Krylov

    User Avatar
    Science Advisor
    Education Advisor

    What do you mean by a simple linearization? And non-zero real part at the fixed point also sounds a bit strange. I would say something like:

    If none of the eigenvalues of the Jacobian matrix associated with the fixed point lie on the imaginary axis, then locally the phase portraits of the non-linear system and its linearization are qualitatively the same.

    The premisse is sufficient, but not necessary: there do exist non-linear systems with a fixed point with associated eigenvalues, some of which purely imaginary, such that locally the phase portraits of the non-linear system and its linearization are still qualitatively the same. (This phenomenon is typical for Hamiltonian systems.)
    It seems a bit of overkill to call this the Jordan normal form, though you are right. The reason I find it overkill, is because here you have a pair of algebraically simple complex eigenvalues, so all Jordan blocks are simply ##1 \times 1##.
    I would say that the stability depend on ##\alpha##. If ##\alpha < 0## then we have stability, if ##\alpha > 0## then we have instability. EDIT: Also, this ##J## is not a Jordan form. (For that it should be upper triangular.) I read it as another example of a possible Jacobian matrix ##A##.
    No, the Jordan normal form theorem also holds for complex ##M##. In fact, if you have eigenvalues with a non-zero imaginary part, then ##M## is going to have some entries with non-zero imaginary part as well.

    However, to determine the type and stability of the fixed point, as you set out to do, you only need the eigenvalues, not the eigenvectors. In particular, in the Linearization Theorem there is no need to "transform back" anything. Just by looking at the eigenvalues, as long as none of them are purely imaginary, you can decide upon the character of the fixed point.
    I find this a good question that is often ignored. Starting from your complex eigenvector corresponding to the complex eigenvalue ##\lambda = \frac{1}{2}(1 + i\sqrt{3})##, you can easily obtain a real-valued general solution of the linearized equation ##\dot{y}(t) = Ay(t)##. It is not my intention to copy standard textbook material here, so let me refer you to Differential Equations, Dynamical Systems & An Introduction to Chaos by Hirsch, Smale & Devaney, 2nd or 3rd edition, 2004 or 2012. Have a look at section 3.2 for a very clear explanation.
     
    Last edited: Nov 30, 2015
  4. Nov 30, 2015 #3

    Krylov

    User Avatar
    Science Advisor
    Education Advisor

    Now it's the end of my day and I have some more time to explain what I meant at the end of my reply. Let ##v = (1,\lambda)## be the complex eigenvector corresponding to ##\lambda##. You can then easily check that ##u(t) :=\Re{[e^{\lambda t}v]}## and ##w(t) := \Im{[e^{\lambda t}v]}## are real-valued, linearly independent solutions of the linear system corresponding to ##A##, implying that the general solution of said system is
    $$
    y (t) = c_1 u(t) + c_2 w(t)
    $$
    for arbitrary real constants ##c_1## and ##c_2##, and this is then of course real-valued as well. Because you already know the eigenvalue ##\lambda##, you can conclude that the motion is circularly outward (i.e. an unstable focus). The lines spanned by ##u(0)## and ##w(0)## in ##\mathbb{R}^2## are invariant and help you to draw the phase portrait of the linear system (in the original coordinates!), hence an approximation of the phase portrait of the non-linear system near the origin. Incidentally, if
    $$
    M :=
    \begin{bmatrix}
    u(0)& w(0)
    \end{bmatrix}
    =
    \begin{bmatrix}
    1& 0\\
    \frac{1}{2}& \frac{1}{2}\sqrt{3}
    \end{bmatrix}
    $$
    then the matrix obtained as
    $$
    M^{-1}AM = \frac{1}{2}
    \begin{bmatrix}
    1& \sqrt{3}\\
    -\sqrt{3}& 1
    \end{bmatrix}
    $$
    is exactly of the form that you said you are already familiar with. This matrix is sometimes called the real Jordan form of ##A##, and unlike the complex case, the real form is not necessarily upper triangular. As you can see from your own question, it is important to keep the real and complex Jordan forms apart.

    In any case, I would recommend you have a look at Chapter 3 of the reference I gave above. There you will find a discussion that goes into much more detail than is possible in a forum post.
     
    Last edited: Nov 30, 2015
  5. Dec 3, 2015 #4

    Krylov

    User Avatar
    Science Advisor
    Education Advisor

    So, did it help? Any thoughts about the above?
     
  6. Dec 3, 2015 #5
    This is the definition that my lecturer gave us. By non-zero real part, she means it may not be a centre, and by simple linearization I think she means the jacobian has to be non-zero. Your definition is much clearer. But as I understand, it says you cannot use the linearization theorem on a focus, for example. But in your second post you use the linearization theorem to approximate a focus? Or am I misunderstanding... ?

    My problem was that my lecturer classified the nature of any fixed point based on how it diagonalizes (or its Jordan form, if it isn't diagonalizable). And a diagonalmatrix with complex entries was not on the list of possible forms, so I didn't know how to classify the point. Part of the problem is that we have never done anything to do with the jordan normal form, so I am unfamiliar with all the technical lingo. She just gave us the nature and the corresponding matrix.

    Yes that's my fault (and partly my lecturer's). All we were told is that if the jordan normal form looks like this ##J=\begin{bmatrix}\alpha&&-\beta\\\beta&&\alpha\end{bmatrix}## then it is a centre if ##\alpha=0##, a stable focus if ##\alpha<0## and an unstable focus if ##\alpha>0##. We were not told that this isn't even in the jordan normal form, nor were we told how to get it into the "real jordan form" .


    Thanks for the tips. I have redone the question using this. (-1,1) is a saddle (I did have to transform the separatrices of this one). (0,0) is an unstable focus and (-1,1) is a stable focus. I also found some isoclines for when the gradient is infinite. These are along the lines y=1,0,-1. All of this information helped me to get a good idea of the phase portrait as a whole. I did find some isoclines where we have constant gradient 0, but these were too hard to plot free hand and I think I have a good enough phase portrait.

    Just out of curiousity, and if you have the time, how would you go about it if they were purely imaginary?


    Thank you for the reference text and showing me how to get it into the real jordan form. I am considering buying this text book!

    Cheers.
     
  7. Dec 7, 2015 #6

    Krylov

    User Avatar
    Science Advisor
    Education Advisor

    It is sufficient to demand that there are no eigenvalues on the imaginary axis. In case of a center, all the eigenvalues are purely imaginary, so indeed the linearization theorem does not apply. However, when all the eigenvalues are of the form ##\lambda_{\pm} = a \pm b i## and ##a \neq 0## (indicating a focus) then they are not purely imaginary and the linearization theorem applies just fine.
    Your confusion is understandable and your question completely justified. When you are dealing with complex eigenvalues, you need to be careful to base your dynamical classification on the real Jordan form, as I explained in my second post. In this context, it is very important to keep the real and complex Jordan forms apart, because indeed: if you use the complex Jordan form (which is the more common of the two) then your resulting matrix is going to be complex and you will not be able to classify it.
    I think that our discussion so far has clarified this?
    If they are purely imaginary, then in general you cannot conclude anything about the non-linear system on the basis of the linearization. (There are exceptions: For example, when the system is Hamiltonian, then "linear centers" correctly predict "non-linear centers".) In this case, a correct approximation of the non-linear system near the fixed point must include higher order terms (quadratic, cubic etc.) A systematic way to do this is provided by "bifurcation theory", and in particular the theory of local normal forms. Bifurcation theory studies, among other things, systems that depend on parameters. It is useful to answer questions such as: If the eigenvalues are ##\lambda_{\pm} = a \pm b i## and ##a \in \mathbb{R}## is considered a parameter, what happens when ##a## passes through zero? (In this case, one of the scenarios is that of the "birth" of a period solution of the non-linear system.) An introduction by examples can be found in Hirsch, Smale and Devaney. Should you need more references, let me know.
    Nice to hear! I think you may like it a lot and it will be very useful.

    The 2nd and 3rd edition don't differ much, is my feeling. They are precise, but more example and application driven than the first edition from 1974 by the first two authors, which goes by the title Differential Equations, Dynamical Systems, and Linear Algebra. This older book is heavier on linear algebra (for example, it includes a general proof of reduction to (real and complex) Jordan canonical form.) It depends on your preferences and learning style, but probably it's nicest to start with the 2nd or 3rd edition and then consult the 1st edition to satisfy any appetite that you may have for mathematical generalization. All editions are beautiful books.
     
    Last edited: Dec 7, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Classify the fixed points of this dynamical system
Loading...