5
$\begingroup$

The following nonlinear system has a fixed point in the origin. I want to know if this fixed point is stable:

\begin{align*} &\dot{\alpha}=\alpha^2-2\beta(\alpha+\beta)\\ &\dot{\beta}=\beta^2-2\alpha(\alpha+\beta) \end{align*}

Note that the linearization of the system leads to $\dot{\alpha}=\dot{\beta}=0$ to first order, so this method is useless. To find out the stability, I wrote this equations as:

\begin{align*} \dot{\alpha}=\left(\alpha,\beta\right) \cdot A\cdot \left(\begin{array}{c} \alpha\\ \beta \end{array}\right),\quad A=\left(\begin{array}{cc} 1 & -1\\ -1 & -2 \end{array}\right) \end{align*}

\begin{align*} \dot{\beta}=\left(\alpha,\beta\right) \cdot B\cdot \left(\begin{array}{c} \alpha\\ \beta \end{array}\right),\quad B=\left(\begin{array}{cc} -2 & -1\\ -1 & 1 \end{array}\right) \end{align*}

Since $A$ and $B$ are real and symmetric, they are diagonalizable, with real eigenvalues. It turns out that $A$ and $B$ have one eigenvalue positive and the other negative (in fact, $A$ and $B$ have the same eigenvalues), explicitly:

\begin{align*} &\text{Eigenvalues of $A$:}\quad \frac{1}{2} \left(-1-\sqrt{13}\right),\frac{1}{2} \left(\sqrt{13}-1\right)\\ &\text{Eigenvalues of $B$:}\quad \text{same of $A$} \end{align*} If, for example, the eigenvalues of $A$ were both positive, then $\dot{\alpha}$ would be positive for all points except origin, and then the origin would be unstable (because trajectories starting out of the origin would increase their $\alpha$'s to infinity). But when both $\dot{\alpha}$ and $\dot{\beta}$ are positive or negative depending on the point $(\alpha,\beta)$, as in this case, how can you determine the stability of the fixed point?

EDIT: The phase portrait is:

enter image description here

$\endgroup$
  • $\begingroup$ What does the phase portrait look like? $\endgroup$ – Moo Jun 25 '17 at 23:24
  • $\begingroup$ Have you tried to do a change of basis so that at least one of the equations has some nice form (such that $\dot y = y^2$ or something)? I think considering $\frac{d}{dt}(\alpha+\beta)$ or similar might lead to a solution of some kind. $\endgroup$ – Daniel Robert-Nicoud Jun 25 '17 at 23:41
  • $\begingroup$ @DanielRobert-Nicoud $\frac{d}{dt}(\alpha+\beta)=\alpha^2+\beta^2-2(\alpha+\beta)^2$, but I can't simplify $\alpha^2+\beta^2$, I also tried with other linear combinations such as $\dot{\alpha}-\dot{\beta}=3(\alpha^2-\beta^2)$. I think that the optimal change of basis would be the diagonalization of $A$ or $B$ $\endgroup$ – adiselann Jun 25 '17 at 23:50
  • 1
    $\begingroup$ Yes, that's the idea. But look at your phase portrait: the straight lines hint at a possible change of basis (that's why I was thinking of $\alpha+\beta$, but it's probably a bit off, as you say). $\endgroup$ – Daniel Robert-Nicoud Jun 26 '17 at 0:04
  • 1
    $\begingroup$ Yes, that solution comes from the reflection symmetry $(\alpha,\beta)\mapsto (\beta,\alpha)$ of your differential equations. That is, if $(\alpha,\beta)=(u(t),v(t))$ is any solution, then $(\alpha,\beta)=(v(t),u(t))$ is also a solution. It follows (from the theory of quotient manifolds) that a solution exists which is left unchanged under the group action. Indeed, the solution with $\alpha=\beta$, i.e. $(\alpha,\beta)=(u(t),u(t))$ is left unchanged under $(\alpha,\beta)\mapsto(\beta,\alpha)$. $\endgroup$ – user254433 Jun 26 '17 at 1:43
2
$\begingroup$

This is mostly a recap of the observations made in the comments, plus some more analysis, because I think it's a nice problem to analyse.

First, both the functional form of the system and the reflection symmetry (see also the phase plane) suggest it's a good idea to introduce $x = \alpha+\beta$, $y = \alpha-\beta$, to obtain \begin{align} \dot{x} &= \frac{1}{2}(y^2 - 3 x^2), \tag{1a}\\ \dot{y} &= 3 x y.\tag{1b} \end{align} We can simplify system (1) somewhat by rescaling $y \to \sqrt{3} y$ and $t \to \frac{1}{3} t$, yielding \begin{align} \dot{x} &= \frac{1}{2}(y^2 - x^2), \tag{2a}\\ \dot{y} &= x y.\tag{2b} \end{align} The phase plane of system (2) looks like this:

enter image description here

What's immediately obvious, is that this phase plane is highly symmetric. It seems to be invariant under rotation over an angle of $\frac{2 \pi}{3}$ (= 120 deg), and rotation over an angle of $\frac{2 \pi}{6}$ seems to keep the shape of the orbits invariant, but changes their flow direction. You can check that both of these observations are indeed correct by considering \begin{equation} \begin{pmatrix} \xi_1 \\ \eta_1 \end{pmatrix} := R(\frac{2\pi}{3}) \begin{pmatrix} x \\ y \end{pmatrix}\quad \text{and} \quad \begin{pmatrix} \xi_2 \\ \eta_2 \end{pmatrix} := R(\frac{2\pi}{6}) \begin{pmatrix} x \\ y \end{pmatrix}, \end{equation} with \begin{equation} R(\theta) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{equation} is the matrix that rotates a vector about the origin over angle $\theta$. Substituting $\xi{1,2}$ and $\eta_{1,2}$ in system (2), you obtain \begin{align} \dot{\xi_1} &= \frac{1}{2}(\eta_1^2 - \xi_1^2), \\ \dot{\eta_1} &= \xi_1 \eta_1, \end{align} and \begin{align} \dot{\xi_2} &= -\frac{1}{2}(\eta_2^2 - \xi_2^2), \\ \dot{\eta_2} &= -\xi_2 \eta_2, \end{align} which implies the above observations.

It seems a good idea to 'factor out' this rotational symmetry present in system (2). Introducing polar coordinates $x = r \cos \theta$, $y = r \sin \theta$, we obtain the dynamical system \begin{align} \dot{r} &= -\frac{1}{2} r^2 \cos(3\theta),\tag{3a}\\ \dot{\theta} &= \frac{1}{2} r \sin(3\theta),\tag{3b} \end{align} which is indeed invariant under $\theta \to \theta + \frac{2 \pi}{3}$. Now, we rewrite the above in terms of the new angle $\phi := 3 \theta$ to obtain \begin{align} \dot{r} &= -\frac{1}{2} r^2 \cos(\phi), \tag{4a}\\ \dot{\phi} &= \frac{3}{2} r \sin(\phi). \tag{4b} \end{align} Now, the nice thing is that we can reinterpret system (4) as being the `polar representation' of some Cartesian system. That is, if we introduce $X$ and $Y$ by $X := r \cos \phi$ and $Y := r \sin \phi$, we can rewrite system (4) in terms of $X$ and $Y$ to obtain \begin{align} \dot{X} &= -\frac{1}{2}(X^2 + 3 Y^2), \tag{5a}\\ \dot{Y} &= X Y. \tag{5b} \end{align} The phase plane of system (5) looks as we would have expected:

enter image description here

Moreover, system (5) turns out to be Hamiltonian, i.e. of the form \begin{align} \dot{X} &= \frac{\partial H}{\partial Y},\\ \dot{Y} &= -\frac{\partial H}{\partial X}, \end{align} with \begin{equation} H(X,Y) = -\frac{1}{2} Y(X^2 + Y^2). \end{equation} Therefore, orbits lie on level sets of $H$, i.e. those curves where $H = E$ (= constant), which can be used to express $X$ in terms of $Y$ as \begin{equation} X = \pm \sqrt{-\frac{2 E + Y^3}{Y}}. \end{equation}

As a final remark, the instability of the origin can be derived from system (5) as follows. Consider the horizontal line $\ell = \left\{ (X,Y) \vert Y=0 \right\}$, i.e. the $X$-axis. Substituting $Y=0$ in system (5), we obtain $\dot{Y} = 0$; therefore, the line $\ell$ is invariant under the flow. In other words, every point on $\ell$ stays on $\ell$ for all time.

It so happens that the origin $(0,0)$ lies on the line $\ell$. Taking a point $(-\epsilon,0) \in \ell$ arbitrarily close to (and directly to the left of) the origin, we see that this point will flow away from the origin, because the flow on $\ell$ is given by $\dot{X} = -\frac{1}{2} X^2$. This holds for all $\epsilon > 0$; hence, we have found an unstable flow direction at the origin, spanned by $(-1,0)$. Therefore, the origin is a (nonlinearly) unstable fixed point of system (5), and in extension of system (1).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.