0

Problem :

Find the Stationary Points and discuss their behavior using Linearisation for the following system of differential equations :

$$x' = \sin y, y' = -\sin x$$

Discussion :

Solving the system :

$$\begin{cases} \sin y = 0 \\ -\sin x =0 \end{cases}$$

yields :

$$\begin{cases} x= k\pi, k\in \mathbb Z \\ y=h\pi, h\in \mathbb Z \end{cases}$$

So, every point $A_k = (k\pi,h\pi)$ is a stationary point of the initial system of differential equations.

Now, the Jacobian of the given system is :

$$J(x,y) = \begin{bmatrix} 0 & \cos y \\ -\cos x & 0\end{bmatrix}$$

The Jacobian of any given stationary point that we found, calculates as :

$$J(k\pi, k\pi) = \begin{bmatrix} 0 & \cos (k\pi) \\ -\cos (h\pi) & 0\end{bmatrix} $$

It's clear that $\det(J(k\pi, k\pi)) = \cos^2(k\pi) \neq 0 $

and that the eigenvalues of the stationary point Jacobian is :

$$\det(J(k\pi, k\pi) - λI) = 0 \Leftrightarrow λ^2 + \cos(k\pi)\cos(h\pi) = \begin{cases} λ^2 +1 = 0 \Leftrightarrow λ = \pm i \\ λ^2 - 1 = 0 \Leftrightarrow λ = \pm 1 \end{cases}$$

So, since the eigenvalues are purely imaginary, according to notes, this tells us that the stationary points $A_k = (k\pi,h\pi)$ are verified as centers of our differential equation system if the first case holds, or as a different kind of critical point ( I do not know it's name in English), which is unstable.

My question is, am I correct ? I found a theorem in our book stating that :

Let $(ξ,n)$ be a hyperbolic stationary point (the Jacobian at the point $(ξ,n)$ is not equal to $0$) of the almost linear system : $$x'=f(x,y), y' = g(x,y).$$ Then, in a neighborhood of $(ξ,n)$ the almost linear system and the linearised system of it, have topologically equal vector fields, which means that thay have the same kind of stability, except of the case when $(ξ,n)$ is a center of the linearised system.

The last part of this theorem (except of the case when $(ξ,n)$ is a center of the linearised system) got me wondering if my solution and hence my conclusion about the stationary points and their behavior, of the given system of differential equations is correct.

I would really appreciate if anyone could clarify me the above and help me get a better grasp of understanding how the theorem correlates with such cases of systems.

Community
  • 1
Rebellos
  • 21,324
  • 2
    The right stationary points are $(k\pi, h\pi)$ with $k,h\in\mathbb Z$. For example, $(0,\pi)$ is a stationary point. If you write $(k\pi, k\pi)$ you miss that. – Giuseppe Negro Nov 15 '17 at 11:15
  • @GiuseppeNegro That's true ! But still, my question holds for what happens regarding the behavior of the stationary points, since the eigenvalues still remain the same. – Rebellos Nov 15 '17 at 11:19
  • 3
    It's as if the nonlinear terms perturb the eigenvalues. Perturb purely imaginary eigenvalues a tiny bit, they're no longer purely imaginary, and you get a focus instead of a center. – Gerry Myerson Nov 15 '17 at 11:21
  • @GerryMyerson I modified the question to hold the corrected case of the stationary points. This means that now, eigenvalues can take 2 different case values, either real and non-equal with their product being less than zero or imaginary. Going over the imaginary ones, is it correct to say that in this particular problem, the stationary point will be a focus instead of a center ? – Rebellos Nov 15 '17 at 11:25
  • I don't know. Have you no study material that discusses this situation? – Gerry Myerson Nov 15 '17 at 11:26
  • @GerryMyerson Only the theorem given. Actually, my question is, is the Jacobian and eigenvalue part discussed over a Linearised system of the almost linear system given, or there is no perturbation on the conclusions ? – Rebellos Nov 15 '17 at 11:28
  • 2
    I don't understand your question. But I'd suggest you find a text that covers this sort of thing, or ask your teacher if this is from a course you're doing. – Gerry Myerson Nov 15 '17 at 11:31
  • @GerryMyerson There is one example of a system involving 4 equations that works over imaginary eigenvalues and shows that it's finally a focus instead of a center, but it gets way more complicated since it involves polar transformations and limit forms. I'll try working around with that though ! I'll make sure I'll ask our professor. Thanks for the comment regarding perturbation of the imaginary eigenvalues, gave me a slight insight on the theorem's case ! – Rebellos Nov 15 '17 at 11:32
  • 1
    Indeed, in full generality, a saddle of the linearized system is a saddle of the nonlinear system and a center of the linearized system is a center or a stable focus or an unstable focus of the nonlinear system. In the present case, every center corresponds to a center because the system follows the level lines of the functional $$V(x,y)=\cos x+\cos y$$ The advice to refer to some source, textbook or otherwise, covering these topics, already formulated here by another user and elsewhere about previous similarly uninformed questions, but neglected, until now, by the OP, seems most relevant. – Did Nov 15 '17 at 11:47
  • If you want a simpler 2D example for non-hyperbolic focus, take a look here. You can modify the second example in such way that the non-hyperbolic focus can be even unstable. – Evgeny Nov 15 '17 at 17:04

0 Answers0