Stability of a system of DE's

1. Jan 11, 2012

fluidistic

1. The problem statement, all variables and given/known data
Find the stationary solution(s) of the following system of DE and determine its stability:
$x'=x^2+y^2+1$.
$y'=2xy$.

2. Relevant equations
$x'=\frac{dx}{dt}=0$, $y'=\frac{dy}{dt}=0$.

3. The attempt at a solution
I tried to google "stationary solutions of a system of DE" but didn't find anything that can help me. I'm guessing they mean solutions that does not change with respect to time, hence $x'=y'=0$. By setting this constraint, I reached that $(x,y)=(\pm i \sqrt {1+y^2},0)=(\pm i , 0)$ are critical points. Namely $(x,y)=(i,0)$ and $(x,y)=(-i,0)$ are 2 critical points, or stationary solutions to the system of DE.
Now I do not know about the stability of such a system. What should I check for?

2. Jan 11, 2012

fzero

To check stability, expand $(x,y)$ around the critical points $(x_0,y_0)$, i.e., let

$$x=x_0 + \alpha,~~~y=y_0 + \beta.$$

Keeping terms that are linear in $(\alpha,\beta)$ and their derivatives lead to the linearization of the original system of ODEs. A critical point is stable if the solutions to the linearized equations with an initial condition in a small neighborhood of the critical point converge uniformly to the critical point at later times. In terms of $(\alpha,\beta)$, both of $(\alpha,\beta)$ will remain small and tend to zero as the system evolves in time. For unstable critical points, solutions to the linear equations tend to take the system away from the critical point.

3. Jan 11, 2012

fluidistic

Ok thank you fzero. I'm not understanding well how to "plug" the uniform convergence here since I don't see a sequence of functions.
What I understand is, if I can show $|x(t)- \tilde x (t)|<\varepsilon$, $\forall t \geq 0$ and $\forall \varepsilon >0$ and then doing the same with y(t), then the point is stable. Maybe the "for all t's" implies the uniform convergence.
Oh and $\tilde x(t)=x_0+\alpha$.
But for this, I'd have to know exactly x(t). Am I missing something, misunderstanding something?

Edit: I think I'm totally wrong, and I'm very tired. Will try to think more about it.

4. Jan 11, 2012

fzero

Yes, you can be more precise about how to define uniform convergence. It is the statement that, for any $\epsilon>0$, there is a $\delta>0$, such that for all $x(t)$, $|x(0)-x_0|<\delta$ implies that $|x(t)-x_0|<\epsilon$ for all $t>0$. What is usually called "stability" here is more precisely called asymptotic stability, where $|x(t)-x_0|\rightarrow 0$ as $t\rightarrow 0.$ (The $x(t)$ for different times is the sequence.)

There is a theorem (associated to Lyapunov) that states that a critical point is asymptotically stable if the eigenvalues of the linearized system have a negative real part. In practice this means that the solutions to the linearized system approach zero exponentially as $t\rightarrow 0$. It is not necessary to find the full solution for $x(t)$.

I don't have an ode book handy, but I'd think that a decent one would have a watered down but more complete explanation. Google will turn up various course notes that might be useful.

Last edited: Jan 11, 2012
5. Jan 12, 2012

fluidistic

Hey fzero, I'm trying to understand more, but not progressing much despite having downloaded and try to understand several lectures/courses on the Internet.
Could you tell me more about the linearization of the system of DE? You were talking about keeping the terms linear in alpha and beta, what did you mean exactly?

6. Jan 12, 2012

fzero

We're interested in the behavior of solutions that start out (have initial conditions) in a very small neighborhood of the critical point. So we can assume that, at least for small times, $\alpha = x - x_0$ is small. It therefore makes sense to expand our DE system in a Taylor series around $x_0$ and, in particular, to approximate the behavior of the system by the lowest order terms. The idea regarding stability is that if the behavior of the linearized system results in convergence of solutions back to the critical point, the nonlinear terms (which start out very small) should not spoil the convergence.

Now I'm not sure if you still have a question about which terms to keep in the expansion. You keep all linear terms in the unknowns and their derivatives, so, for example

$x^2 y' \sim x_0^2 \beta', ~~~xy^2 \sim x_0y_0^2 + y_0^2 \alpha + 2x_0y_0 \beta.$

7. Jan 12, 2012

fluidistic

Ok thanks a lot, this is clearer now in my mind, however I cannot figure out where did that mathematic expression come from? From the system of DE's...? Hmm how?

8. Jan 12, 2012

fzero

I purposely picked expressions that didn't appear in your system of equations (out of habit since this is a HW forum). If you can fill in the steps for those, you'd have no problem working out the system in your OP.

9. Jan 12, 2012

fluidistic

Ah nice examples then, because clearly I'm again missing something.
For $xy^2$, here are my thoughts: Taking $x=x_0+\alpha$ and $y=y_0+\beta$, I reach that it's worth $x_0y_0^2+2x_0y_0 \beta +2 \beta \alpha y_0 + \alpha y_0 ^2+ \beta \alpha ^2+ x_0 \beta ^2$. Taking the terms that are linear in $\alpha$ and $\beta$ and their derivatives I reach what you found: $x_0 y_0 ^2 +2x_0 y_0 \beta +\alpha y_0 ^2$.
However when I do the same reasoning for $x_0^2y_0 '$ I get that it's worth $x_0 ^2 y_0 '+2x_0y_0'+\beta ' x_0^2+2x_0 \beta '$ and I all the terms are linear in terms of $\alpha$, $\beta$ and their derivatives so I cannot neglect any term here. I have in fact 3 extra terms compared to your result, I do not understand why.

10. Jan 12, 2012

fzero

You should find

$$x^2y '=x_0 ^2 y_0 '+2x_0y_0'+\beta ' x_0^2+2x_0\alpha \beta '.$$

The first two terms vanish because $y_0'=0$ (the values at the critical point are constants). The last term, proportional to $\alpha \beta '$, is considered to be quadratic in the perturbations and so we neglect it when we linearize.

There's probably a subtle point in assuming that the derivatives of the small perturbations aren't too large themselves. I'd expect that if they were going to be large in the vicinity of the critical point, the linearization test would not indicate stability anyway, but I'm not sure of a detailed argument for that.

11. Jan 13, 2012

fluidistic

Ah perfect.
I get as linearization of the system of ODE's:
$\alpha ' = x_0 ^2 +2x_0 \alpha + y_0 ^2 +2y_0 \beta +1$.
$\beta '=2x_0y_0+2x_0 \beta +2 \alpha y_0$.
With $(x_0, y_0)=(i,0)$ I reach $\alpha =Ae^{2it}$ and $\beta =Be^{2i t}$. Both diverge when t grows up so the critical point is a saddle point?
For the point $(-i,0)$ at first glance it looks like I'd reach $\beta = Ce ^{-2it}$ and $\alpha = D e^{-2it}$ wich converges to 0 when t goes to infinity and remains bounded for small t's. But I guess I must use the uniform convergence here.

Edit: Discard what I just said, I didn't realize the exponential was complex...

Last edited: Jan 13, 2012
12. Jan 14, 2012

fzero

I hadn't been paying attention to the particular details of this system and it turns out that this is a special case. First, I'll point out that due to the symmetries of fixed point equations, since $(\pm i, 0)$ are fixed points, so are $(0,\pm i)$.

Second, as to the nature of the fixed point, the behavior of the solutions is clearly not asymptotically stable, since we don't have convergence back to the fixed point. However, the system isn't unstable either. The linearized solutions oscillate around the fixed point tracing out an ellipse whose diameters are set by the normalizations. This case is called stable and the fixed point is called a "center point."

As I said this is a special, or at least individual, case that is distinct from the asymptotically stable and unstable cases. It's kind of a bizarre first example to tackle when learning the material, especially because the solutions are genuinely complex, whereas if they were real (sin or cos) we could at least straightforwardly plot them to see how this looks. Obviously by a redefinition of variables we can put it into a form where the solutions are sin and cos, so you might want to do a quick sketch in the case that $\tilde{x} = A \cos t,~\tilde{y}=B\sin t$ to see that the picture I described above makes sense.

13. Jan 14, 2012

fluidistic

Ok thank you very much for all your time.
Instead of drawing x tilde and y tilde, is the following correct?
x(t) and y(t) correspond to the trajectory of a particle M. The critical points correspond to the center of mass of a system of 2 particles in which M is one of them. If we set the particle M originally (t=0) close to a critical point (x0+alpha and y0+beta) then the trajectory of the particle M will be an ellipse (that in fact occur in the complex plane). The motion is stable in the sense that the particle will never "go to infinity" but isn't asymptotically stable either because the particle will never reach a critical point (unless it starts there at t=0).

14. Jan 16, 2012

fzero

I don't really like this analogy, since there's nothing in the dynamics that connects it to a system of particles (for example, the interaction law is not specified). It is true that if this system did describe the motion of a particle, then it would be as you describe in the region of the critical point. It's probably best to not try to read too much beyond this system being the rules that describe the evolution of a pair of functions.