# Stable manifold for bidimensional nonlinear dynamic system with complex eigenvalues

Given a autonomous nonlinear dynamic system of the form

$$f(x,y)=\begin{bmatrix} B_1 x + g_1(x,y) \\ B_2 y + g_2(x,y) \end{bmatrix}$$

with $B_i\in\mathbb{R}$ (bidimensional problem), with $g_i$ "purely" non-linear in $x$ and $y$, solve the equation

$$\dot{X} = f(X), \ X=[x,y]^T$$

is a classical problem in dynamics (with $f,g_i$ smooth enough) in the neighbourhood of a hyperbolic fixed point. The stable manifold theorem (together with the implicit function theorem, I think) yields that there exists a function which describes all orbits near the fixed point.

1- Is all this correct?

I am trying to be exhaustive and solve examples of all possibles such cases, that is with different kind of hyperbolic points.

For the following cases, it has been straightforward: [both strictly positive (negative) eigenvalues], [one eigenvalue strictly positive, one eigenvalue strictly negative], [equal real non-zero eigenvalues].

The remaining cases involve complex (conjugate) eigenvalues, where I am stuck.

2- When I observe a stability diagram (in the phase plane) of a hyperbolic saddle point for example, the "shape" of the stable manifold appears to be quite obvious. For the complex cases, the theorem states that the stable manifold is of dimension 2, but I can hardly understand this: what does it look like? This question also hold for the linear case and the stable subspace.

3- It does not seem possible to write the equations in the form of the first equation of the post for complex (non-real) eigenvalues, since either $B_i$ are not real, either, when written as a real Jacobian block, both $x$ and $y$ appears in both lines. The question, closely related to question 2, is: how can the stable manifold be calculated when the eigenvalues are (non-real) complex?

Edit

4- Additional question: why is it that the vector spaces in which $x$ and $y$ live have to coincide with the dimension of the stable and respectively unstable eigenspaces?

Edit: answer to question 4: the question seems to be not very clear. I was wondering why in the first equation the number of $B_1$ lines had to coincide with the dimension of the stable manifold, and why the number of lines of $B_2$ had to be the dimension of the instable manifold. I now understood that this is not necessary. It is commonly done because people are usually looking for the equation of the stable manifold so seek it as a curve $(x,s(x))$. But any invariant manifold can be calculated by splitting $X$ in another way, therefore changing the dimensions of $B_1$ and $B_2$.

First I'd like to mention that you can see a hyperbolic point as the transversal intersection of two invariant manifolds, e.g., the stable $(W^s)$ and the unstable ($W^u$) invariant manifold. Since the intersection is transversal, the direct sum of their respective tangent spaces is of dimension 2. Therefore you can identify it with $\mathbb R^2$. The previous description goes in the direction of your question 4. In dimension $2$ the only way you can intersect transversally two $1$-dimensional manifolds is in such a way, that their tangent spaces are not tangent, see figure below.

For your other question, the difficulty is of course the complex nature of the eigenvalues, but you can qualitatively think as: "whenever there are complex eigenvalues, the orbits spin around". Let us take a look on the Jacobian matrix at zero. For brevity let

$$J|_{(0,0)}=\begin{bmatrix} a_1 & a_2\\ a_3 & a_4 \end{bmatrix}$$

The characteristic polynomial is $$p(s)=s^2-\tau s+\delta,$$ where $\tau=(a_1+a_4)$ denotes the trace of $J$ and $\delta=a_1a_4-a_2a_3$ denotes the determinant of $J$. The eigenvalues are, as you know, $$\lambda_{1,2}=\frac{\tau\pm\sqrt{\tau^2-4\delta}}{2}.$$

For the complex conjugated case then, you have the condition $\tau^2-4\delta<0$. Observe that you can characterise all the behaviour in the following diagram of $(\delta,\tau)$.

That is, you are now interested in what happens inside the parabola $\tau^2-4\delta=0$. You can easily show that if you are inside the parabola, the solutions (of the linearised problem) can be written as

$$x(t)=\exp(at)(A\cos(bt)+B\sin(bt))$$ $$y(t)=\exp(at)(C\cos(bt)+D\sin(bt))$$

Explaining the stability and instability of the orbits, as shown in the next diagram.

To conclude this part, I should bring to your attention Hartman-Grobman theorem. Which in plain words states two important things for systems near a hyperbolic equilibrium point: 1) Two systems with the same type of eigenvalues are topologically equivalent. In particular nodes are equivalent to stable foci. 2) A nonlinear system is equivalent to its linearisation.

Ok, now that you have some geometric insight, how do you see the invariant manifolds in the complex case? Well when I learned the topic, the following helped me understand things. Consider the linear problem

$$\dot X=\begin{bmatrix} \dot x\\ \dot y \end{bmatrix}=\begin{bmatrix} a & -1\\ 1 & a \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$$

Note that the eigenvalues are: $\lambda_{1,2}=a\pm j$. Let $x=r\cos\theta$, $y=r\sin\theta$, then you uncouple the equations obtaining:

$$\dot r=ar, \, \dot\theta=1.$$

You now study the one dimensional equation $\dot r=ar$, where you already know how to visualise the invariant (stable or unstable) manifold.

• Thank you for this very nice answer. I think I was confused by the fact that there are several types of stable manifolds (stable if $a<0$ in your example): the unidmensional ones, which are every orbits, and their union, which is $\mathbb{R}^2$. When the eigenvalues are real, you can "see" the stable manifold tangent to the origin. In the complex case, their "do not appear". In other words, in the linear case, the eigenvectors are obvious (straight lines) when you draw several orbits, while if the eigenvalues are complex, no such line appear. Is that it? (I'll comment on other points later!) – anderstood Aug 4 '14 at 9:29
• You are correct in your idea. Imagine a node. Every orbit is an invariant manifold. In fact every orbit is an invariant stable manifold. Their union form a 2 dimensional invariant stable manifold. The advantage of having real eigenvectors is that it is enough to study the generated spaces, in the sense that if you know what happens there, then you know what happens everywhere (in the linear case of course). For the complex case, recall that you can see a complex number in the real plane. So a 2d real system with complex eigs. can be seen in a 1d complex system (like passing to polar coords.) – PepeToro Aug 4 '14 at 11:21

For both questions 2 and 3: In the complex case the (un)stable manifold is the space spanned by the vectors $Re(v)$ and $Im(v)$, where $v$ is an eigenvector of $A$ (for the linear case of course). Note that the other eigenvector is $\bar{v}$, which is the complex conjugate of $v$. Also, it is not hard to prove that $Re(v)$ and $Im(v)$ are linearly independent. Detailed information can be found in http://www.math.utk.edu/~freire/complex-eig2005.pdf.

For question 4: First note that if $Av_i=\lambda_i v_i$, then $e^{A}v_i = e^{\lambda_i}v_i$. Now the solution to the linear autonomous system is $x(t)=e^{At}x_0$. Suppose that $x_0=a_1 v_1+\dots+a_n v_n$, then $x(t)=a_1 e^{At}v_1+\dots+a_n e^{At}v_n=a_1 e^{\lambda_it}v_1+\dots+a_n e^{\lambda_it}v_n$. This is why $x(t)$ lives in the space spanned by the eigenvectors of $A$. The stable space is the space spanned by the eigenvectors $v_i$ such that $\text{Re}(\lambda_i)<0$. This means if $x_0 \in W^s$, then $x(t) \in W^s$ for all $t$ and $\lim_{t\to \infty}x(t)=0$.

For the first question I'm not sure what do you mean by a function describing all orbits.