# If the eigenvalues $\lambda_1$ and $\lambda_2$ for a linearized system are equal to $\pm bi$…

What does this suggest about the corresponding phase portrait at some critical point of the system? My textbook says "stable or unstable, center or spiral point," but how could I tell which it really is in any particular circumstance?

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For example, consider the nonlinear system \eqalign{\dot{x} &= -y + \alpha (x^2 + y^2) x\cr \dot{y} &= x + \alpha (x^2 + y^2) y\cr} whose linearization has eigenvalues $\pm i$, and thus a centre, at $(0,0)$. In polar coordinates, \eqalign{\dot{r} &= \alpha r^3\cr \dot{\theta} &= 1\cr} so the system has the solutions \eqalign{r &= (r_0^{-2} - 2 \alpha t)^{-1/2}\cr \theta &= \theta_0 + t\cr} If $\alpha < 0$, the origin is asymptotically stable (all solutions approach $r=0$ as $t \to +\infty$) while if $\alpha > 0$ it is unstable (all solutions approach $r=0$ as $t \to -\infty$). It is only a centre (stable but not asymptotically stable) for the linear case $\alpha = 0$.

In general, things are not so neat. You can try to see what happens to $r$ after one revolution around (and close to) the equilibrium. If after one revolution it is always closer to the equilibrium than when it started, that makes the equilibrium asymptotically stable (a stable spiral). If it is always farther, the equilibrium is unstable (an unstable spiral). In the latter case there might be a limit cycle.

EDIT: For example, consider the system \eqalign{\dot{x} &= -y + \alpha x^3\cr \dot{y} &= x + x y\cr} which has the same linearization. In polar coordinates, this becomes \eqalign{\dot{r} &= {r}^{2}\cos \left( \theta \right) \left( 1- \left( \cos \left( \theta \right) \right) ^{2}+\alpha\,r \left( \cos \left( \theta \right) \right) ^{3} \right) \cr \dot{\theta} &= 1-\sin \left( \theta \right) \alpha\,{r}^{2} \left( \cos \left( \theta \right) \right) ^{3}+\sin \left( \theta \right) r \left( \cos \left( \theta \right) \right) ^{2} \cr} We can think of $r$ as a function of $\theta$, with $$\dfrac{dr}{d\theta} = \dfrac{\dot{r}}{\dot{\theta}}$$ Now taking initial condition $r(0) = r_0$, write $$r(\theta) = r_0 + a_2(\theta) r_0^2 + a_3(\theta) r_0^3 + \ldots$$ substitute this into the equation for $dr/d\theta$, expand in powers of $r_0$, and solve the resulting, rather complicated, system of differential equations with initial conditions $a_j(0)=0$. We may need several terms until we find one where $a_j(2 \pi) \ne a_j(0)$. In this case the first turns out to be $$a_5(2\pi) = \dfrac{\pi}{64} (9 \alpha^3 - \alpha - 2)$$ Thus when this is positive, the equilibrium is an unstable spiral, and when it is negative, the equilibrium is a stable spiral.

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But this system has purely real eigenvalues, and we were discussing purely imaginary. Is it true that a critical point with purely imaginary eigenvalues is a center (or at least Lyapunov stable)? – Gyu Eun Lee Dec 11 '12 at 5:29
No, this system has purely imaginary eigenvalues (I apologize for typing $\pm 1$ when I meant $\pm i$). – Robert Israel Dec 11 '12 at 5:31
Ah. I worked out the eigenvalues myself, and you're right. And I've found the Hartman-Grobman theorem after some searching. – Gyu Eun Lee Dec 11 '12 at 5:45

In the case that the eigenvalues are purely imaginary, this is a center and hence Lyapunov stable. You only have spiral points (I call them foci) if you have eigenvalues with nonzero imaginary parts. In this case stability depends on the sign of the real part of the eigenvalues (which are complex conjugates, so the real part is equal).

The idea is that your solution behaves like this: if the eigenvalue is $\lambda=a+ib$, then the solution is like $e^{\lambda t}=e^{(a+ib)t}=e^{at}e^{ibt}=e^{at}(\cos bt + i \sin bt)$, where the last part is Euler's formula. The $e^{at}$ part, which comes from the real part, describes exponential behavior of the solution. So if $a > 0$, the magnitude of the solution becomes arbitrarily large as $t \to \infty$, since $e^{at}\to\infty$. So this gives asymptotic stability. If $a < 0$, then the magnitude becomes arbitrarily small since $e^{at}\to 0$. So this gives instability.

If $a=0$ and $b\neq 0$, then our solution behaves like $(\cos bt + i \sin bt)$, which is periodic and in fact traces a circle.

If $a \neq 0$ and $b=0$, then the solution doesn't spiral at all, since $(\cos bt + i \sin bt)=1$. In this case the solution looks like exponential curves.

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This is for a linear system. A nonlinear system with pure imaginary eigenvalues could have a stable or unstable spiral. – Robert Israel Dec 11 '12 at 4:32
This will work locally, since the OP says the system is linearized. – Gyu Eun Lee Dec 11 '12 at 4:33
The linearized system has a centre, but the nonlinear system that has this linearized system might not. – Robert Israel Dec 11 '12 at 4:53
I'm not sure what exactly you're saying. As far as I know the system should have a center if you go sufficiently close to the critical point where the eigenvalues are purely imaginary. Of course the system may have other critical points that aren't centers, but this critical point should be a center. Maybe you could provide an example? We might just be missing each other. – Gyu Eun Lee Dec 11 '12 at 5:03
I've provided an example in my answer. – Robert Israel Dec 11 '12 at 5:21