6
$\begingroup$

Given a system of differential equations \begin{eqnarray} x'&=&2y(z-1)\\ y'&=&-x(z-1) \quad (1)\\ z'&=&xy \end{eqnarray} Note that $u_0$=(0,0,0) is an equilibrium point of the system. Let $V(x,y,z)=x^2+2y^2$ is a Lyapunov function for (0,0,0). Since $V(u_0)=0$, $V(u)> 0$ for $u\neq u_0$ and $\frac{d}{dt} V(x,y,z)=0$ for $u\neq u_0$ then $u_0$=(0,0,0) is stable by Lyapunov theorem.

But I don't understand that $u_0$=(0,0,0) is stable, infact it might be unstable. Here is my argument: Let $r(t)=(x(t),y(t),z(t))$ be the flow of (1). Since $\frac{d}{dt} V(x,y,z)=0$ then $r'(t)\cdot \nabla V=0$. Hence, $r(t)$ always be in a surface $V(x,y,z)=x^2+2y^2=C$ (A cylinder). If we make the radius of cylinder smaller then $x$ and $y$ component of $r(t)$ tend to 0 but $z$ component of $r(t)$ can be infinite since the height of cylinder is infinite. So, $r(t)$ can be far away from (0,0,0) and $u_0$=(0,0,0) is unstable.

Where is the error of my argument? Can we guarantee that $z$ component of $r(t)$ can't go to infinity? Can we esplain that $u_0$=(0,0,0) is stable without Lyapunov?

Thanks a lot.

  • 0
    @coffeemath: It is obviously stable, if my answer below is correct.2012-11-24

3 Answers 3

1

At a specific height $z$ the equations $x'=2y(z-1)$ and $y'=-x(z-1)$ (looking at only the $x,y$ coordinates) make the trajectory go around an ellipse $x^2+2y^2=C$. (By this I mean ignoring where the $z$ coordinates are, i.e. looking at the projection of the trajectory onto the $xy$ plane.)

As the curve goes around the ellipse, the signs on $x,y$ go through the usual things in the four quadrants, so that $xy$ is first positive, then $0$, then negative, then $0$ and so on. So maybe this accounts for $z$ not going "off to infinity" by means of $z'=xy$, since because of the alternating signs on $xy$ the $z$ coordinate will oscillate rather than escape to infinity. Of course a better check would be to actually find the trajectories in closed form, but I don't see that easily. Consider this answer as only a possible explanation of why the trajectory might not escape to infinity in the $z$ direction.

  • 0
    How about the possiblity if the solution only trace the first octan? In this octan, z'=xy>0 so z always increasing. What make you sure that the solution trace along all quadrant of domain2012-12-05
2

The solution stays bounded, and $z$ undergoes periodic motion. The reason for this is symmetry: If you replace $x$ by $-x$ and $t$ by $-t$, the system is unchanged. Therefore, while $(x,y)$ travels once around its ellipse $x^2+2y^2=\text{constant}$ from $x=0$, $y>0$ back to the same point, $z$ must have returned to its original value.

I know that $x$, $y$ traverses the entire ellipse because the first two equations are just a variation of $x'=2y$, $y'=-x$, which has that property. More precisely, introduce some new time variable $\tau$. Write $x_\tau$ etc for the derivative wrt $\tau$ and $x_t$ etc for the derivative wrt $t$, then $x_\tau=x_t\tau_t$, so if $\tau$ is defined by $\tau_t=z-1$, then the first two equations become $x_\tau=2y$, $y_\tau=-x$. This works fine so long as $z<1$. But we are interested in stability at the origin, and if we start close to the origin then $z$ has no chance to grow as big as $1$ in the time it takes to traverse the ellipse.

  • 0
    @HaraldHanche-Olsen: I agree that it is obvious that the linear system traverses the ellipse. However, it is not immediate that the original system must follow suit. It does, but it took me some time to establish the fact formally. Hopefully there is a more succinct solution than the one I offered.2012-12-07
2

It is true that $(0,0,0)$ is Lyapunov stable. The general idea has been suggested in other solutions, here are the truly gory details.

Let $p(t) = (x(t), y(t), z(t))$.

We need to establish that a unique solution exists for all time. Let $r>0$ and let $K_r = \overline{B}(0,r)$. Since the system is smooth, it is Lipschitz on any bounded set. Let $p(t_0) \in B(0,\frac{1}{2}r)$ be an initial condition at time $t_0$. By the existence and uniqueness theorem, a solution exists in $K_r$ for some small time interval $[t_0,t_1]$. As above, we notice that $x(t)^2+2 y(t)^2=x_0^2+ y_0^2$, and hence $(x(t),y(t))$ remains bounded. Since $\dot{z} = x y$, we have $|z(t)-z(t_0)| \leq \frac{1}{2}(x_0^2+ y_0^2)(t-t_0)$. Consequently, we see that for any given $t_1$ we can choose $r$ large enough so that $p(t) \in K_r$ for all $t\in [t_0,t_1]$. It follows that the solution exists and is unique for all $t \geq 0$.

Now we will establish that $t \mapsto p(t)$ is periodic. We know that $(x(t),y(t))$ lies on the ellipse $x^2+2y^2 = V_0$, where $V_0 = x_0^2+2 y_0^2$. First we will show that $(x(t),y(t))$ traverses the entire ellipse without $z(t)$ growing 'too much'. Then we will show that $z(t)$ is periodic as well.

We know that $|z(t)| \leq |z(t_0)|+\frac{1}{2}V_0 (t-t_0)$. Hence if $|z(t_0)| < \frac{1}{4}$ and $(t-t_0) < \frac{1}{2 V_0}$, then $|z(t)| < \frac{1}{2}$. In particular, for any $T>0$, if $V_0$ and $|z(t_0)|$ are 'small enough', then $|z(t)| < \frac{1}{2}$ for any $t \in [t_0,t_0+T]$.

Since $(x(t),y(t))$ lie on the ellipse, we can write $x(t) = \sqrt{V_0} \cos \theta(t)$, $y(t) = \sqrt{\frac{v_0}{2}} \sin \theta(t)$ for some $C^1$ function $\theta$. Differentiating gives \begin{eqnarray} \dot{x}(t) &=& -\sqrt{V_0}\sin \theta(t) \dot{\theta}(t) &=& -2 \sqrt{\frac{v_0}{2}} \sin \theta(t) (1-z) \\ \dot{y}(t) &=& \sqrt{\frac{v_0}{2}} \cos \theta(t) \dot{\theta}(t) &=& \sqrt{V_0} \cos \theta(t) (1-z) \end{eqnarray} from which it follows that $\dot{\theta}(t) = \sqrt{2} (1-z)$. Now choose $T = 4 \sqrt{2} \pi$, which gives a neighborhood $U$ of $(0,0,0)$ such that if $p(t_0) \in U$, then $|z(t)| < \frac{1}{2}$ for any $t \in [t_0, t_0+T]$. Then we have $\dot{\theta}(t) \geq \frac{1}{\sqrt{2}}$, and hence $\theta(t_0+T)-\theta(t_0) \geq \frac{1}{\sqrt{2}} T = 4\pi$ (I choose $4 \pi$ so I can find a point where the positive $x$-axis is crossed and still have $2 \pi$ 'left to go'). Hence $(x(t),y(t))$ traverses the entire ellipse in some $[t_0,t_0+\delta] \subset [t_0, t_0+T]$. (It does not yet follow that $(x(t),y(t))$ is periodic.)

Let $\tau_0 \geq t_0$ be the first time at which $(x(t),y(t))$ crosses the positive $x$-axis, let $\tau_1 > \tau_0$ be the first time at which $(x(t),y(t))$ crosses the positive $y$-axis, and let $\Delta = \tau_1 -\tau_0$. For $t \in [\tau_0+\Delta, \tau_0+2\Delta]$, define $\tilde{x}(t) = -x(2\tau_0+2 \Delta-t)$, $\tilde{y}(t) = y(2\tau_0+2 \Delta-t)$ and $\tilde{z}(t) = z(2\tau_0+2 \Delta-t)$. It is straightforward to verify that $\tilde{p} = (\tilde{x},\tilde{y},\tilde{z})$ satisfies the differential equation with initial condition $\tilde{p}(\tau_0+\Delta)=p(\tau_0+\Delta)$. By uniqueness, we have $p(t) = \tilde{p}(t) $ for $t \in [\tau_0+\Delta, \tau_0+2\Delta]$, and hence $(x(\tau_0+2\Delta), y(\tau_0+2\Delta),z(\tau_0+2\Delta))=(-x(\tau_0), y(\tau_0),z(\tau_0))$ (note that $y(\tau_0) = 0$ by choice of $\tau_0$).

Now we repeat the process for the remaining semi-ellipse. For $t \in [\tau_0+2\Delta, \tau_0+4\Delta]$, define $\tilde{x}(t) = x(2\tau_0+4 \Delta-t)$, $\tilde{y}(t) = -y(2\tau_0+4 \Delta-t)$ and $\tilde{z}(t) = z(2\tau_0+4 \Delta-t)$. Repeating the above procedure, we obtain $p(t) = \tilde{p}(t) $ for $t \in [\tau_0+2\Delta, \tau_0+4\Delta]$, and hence $(x(\tau_0+4\Delta), y(\tau_0+4\Delta),z(\tau_0+4\Delta))=(x(\tau_0), -y(\tau_0),z(\tau_0)) = p(t_0)$ (note that $y(\tau_0) = 0$ by choice of $\tau_0$).

It follows by uniqueness that $p(t+4 \Delta) = p(t)$ for all $t\geq t_0$. Furthermore, $|z(t)| \leq |z(t_0)| +\frac{1}{2}V_0 4 \Delta$ for all $t\geq t_0$. Hence for all $\epsilon > 0$, there is a neighborhood $U$ of $(0,0,0)$ such that if $p(t_0) \in U$, then $\|p(t)\| <\epsilon$ for all $t \geq t_0$.