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Consider the system $$x'=(\epsilon x+2y)(z+1)$$ $$y'=(\epsilon y-x)(z+1)$$ $$z'=-z^3$$

(a) Show that the origin is not asymptotically stable when $\epsilon=0.$

(b) Show that when $\epsilon <0,$ the basin of attraction of the origin contains the region $z>-1.$

I did (a) showing that at least one eigenvalue of the jacobian matrix had not negative real part.

And I'm stuck in (b), I don't understand what do I have to prove?

Can someone help please?

Rodrigo de Azevedo
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user441848
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  • You need to show: when $\epsilon$ is negative, initial values $(x_0,y_0,z_0)$ where $z_0>-1$ will approach the origin as $t \rightarrow \infty$. – Alex Jones Dec 21 '17 at 05:35
  • @AlexanderJ93 Will I use (a)? – user441848 Dec 21 '17 at 05:43
  • Most likely not, part (a) deals with $\epsilon=0$ while part (b) deals with $\epsilon<0$. You may want to look at the Jacobian for negative $\epsilon$ values though. – Alex Jones Dec 21 '17 at 05:45
  • @AlexanderJ93 why? to find the stability of the equilibrium points or something? I don't understand your idea. – user441848 Dec 21 '17 at 06:01

4 Answers4

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Hints: Instead of considering $x^2+2y^2+z^2$ it is in fact easier to consider $z$ separately from $(x,y)$.

1) Integrate the equation for $z$ and show that $z(t)$ goes to zero as $t\rightarrow +\infty$.

2) Show that the function: $L(t)=x(t)^2+2y(t)^2$ verifies the ode: $\dot{L}(t) =-2|\epsilon| L(t) \; (z(t)+1).$

3) Integrate this ode for $L$ and show that $L(t)\rightarrow 0$ as $t\rightarrow +\infty$.

(Along the way you should also conclude that neither $L$, nor $z$ can go to infinity)

H. H. Rugh
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  • Yes seems like it's easier this way to see that $L(t)\to 0$. In your comments you said that as $L(t)\to 0$, then every point in the space is in the basin of attraction, but I don't understand why, can you explain this intuitively ? – user441848 Dec 22 '17 at 22:07
  • When $z(0)<-1$ the system is at first expanding in $(x,y)$, but after some time $z(t)$ becomes larger than $-1$ and then the system starts to contract. This is seen in the behavior of $L(t)$, provided you succeed doing part (3) in my hints. Note that as $L(t)$ goes to zero then both $x(t)$ and $y(t)$ are forced to go to zero as well, and $z(t)$ in any case goes to zero. Hope this clarifies. – H. H. Rugh Dec 22 '17 at 23:02
  • I see, I have another question where does the ode $\dot{L}=-2|\epsilon| L (z+1)$ come from? – user441848 Dec 23 '17 at 00:07
  • I understand that the point with your hints is to show $L(t)\to 0$ but how did you up with that method? It's strange :) – user441848 Dec 23 '17 at 00:14
  • The more 'natural' method is indeed to look at a standard Lyapunov function like $M=x(t)^2+2y(t)^2+z(t)^2$. When $z>-1$ and $M>0$ then $\dot{M}<0$ but it is not bounded by $-cM$ (contrary as to what I commented at some point). So it is a bit more delicate to show that it goes to zero (since it could simply be decreasing but converging to a non-zero value). The reason is the $z^2$ part, whose derivative is $-z^4$ (and the forth power is a problem). Anyway, the simplest way is simply to take the $z$ part aside. – H. H. Rugh Dec 23 '17 at 00:27
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I'm quite sure that this is a problem from Hirsch, Smale, and Devaney's book in Dynamical systems. So let's take it piece by piece.

a) When $\epsilon = 0$ we have that

$$x ' = 2y(z+1)$$ $$y' = -x(z+1)$$ $$z' = -z^3$$

Now we want to show that the origin is not asymptotically stable. This means there exists some solution that does not tend towards the origin as $t \to \infty$ but stays close to the origin (being a little imprecise here, but you can make it precise by looking at the definition of stable).

Let's try to construct such a solution. For simplicity, let's make the bold assumption that $z = 0$, in which case we now have the equations

$$x' = 2y$$ $$y' = -x$$ $$z' = 0$$

This is basically a $2 \times 2$ system with matrix

$$\begin{bmatrix} 0 & 2 \\ -1 & 0 \\ \end{bmatrix}$$

which has eigenvalues $\pm i \sqrt{2}$. Thus we have solutions to this system of form

$$\begin{bmatrix} x \\ y \end{bmatrix} = C_1 \begin{bmatrix} \cos{t\sqrt{2}} \\ -\sin{t\sqrt{2}} \end{bmatrix} + C_2 \begin{bmatrix} \sin{t\sqrt{2}} \\ \cos{t\sqrt{2}} \end{bmatrix} $$

Putting it together,

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = C_1 \begin{bmatrix} \cos{t\sqrt{2}} \\ -\sin{t\sqrt{2}} \\ 0\end{bmatrix} + C_2 \begin{bmatrix} \sin{t\sqrt{2}} \\ \cos{t\sqrt{2}} \\ 0\end{bmatrix} $$

which are ellipses around the origin. These are stable but not asymptotically stable.

b) Here, we can use LaSalle's Invariance Principle. On page 195 (section 9.2 of Hirsch, Smale, and Devaney) we saw that

$$L = x^2+2y^2+z^2$$

is a Liapunov function. Further, we saw that it is a strict Liapunov function since

$$\frac{dL}{dt} = \epsilon(x^2+2y^2)(z+1)-z^4 < 0$$

for $\epsilon < 0$ and $z > -1$.

LaSalle's Invariance Principle given in the textbook says:

Let $X^*$ be an equilibrium point for $X'= F(X)$ and let $L: U \to \mathbb{R}$ be a Liapunov function for $X^*$, where $U$ is an open set containing $X^*$. Let $P \subseteq U$ be a neighborhood of $X^*$ that is closed and bounded. Suppose that $P$ is positively invariant, and that there is no entire solution in $P\setminus\{X^*\}$ on which $L$ is constant. Then $X^*$ is asymptotically stable and $P$ is contained in the basin of attraction of $X^*$.

Here's what we will do. Consider the set

$$P_{abc} = \{(x,y,z) \in \mathbb{R}^3 \quad | \quad |x| \le a, |y| \le b, -1+c \le z \le b$$

Notice that

$$\bigcup_\limits{a > 0, b > 0, 0<c<1} P_{abc} = \{(x,y,z) \in \mathbb{R}^3 \quad | z > -1 \}$$

Furthermore, we have that each $P_{abc}$ contains the origin and is closed. Also, since the Liapunuv function is strict, then each $P_{abc}$ is positively invariant else it would imply that the Liapunov function is increasing (WARNING: THIS IS FALSE. As noted by a comment below, it is NOT true that each of these $P_{abc}$ are positively invariant. I am leaving this solution up here so that you can be warned of my error). Furthermore, having a strict Liapunov function also guarantees that $L$ is not constant along a solution. Thus by LaSalle's Invariance Principle, we must have that each $P_{abc}$ is in the basin of the origin and so the union is also in the basin.

carsandpulsars
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    The set $P_{abc}$ is not positively invariant. Consider, for example, the case $\epsilon=-1$, $a=1$, $b=1$. In this case the first equation is $\dot x=(-x+2y)(z+1)$. The solution that starts at $x_0=(1,1,0)\in P_{abc}$ goes out from $P_{abc}$ since $\left.\dot x\right|_{x_0}=1>0$, so $x$ increases along the trajectory – AVK Dec 21 '17 at 08:30
  • I am inclined to agree with you here. I'll edit my answer and put in a warning where I make my mistake.

    Do you know how to resolve my solution? Or does this approach not work. This is actually how I did it for a homework assignment and my instructor I guess didn't catch it.

     – carsandpulsars Dec 21 '17 at 19:01
  • Any solution can intersect the plane $z=-1$ in one direction only: from $z<-1$ to $z>-1$. This follows from the fact that $\left.\dot z\right|_{z=-1}=1>0$ i.e. $z$ increases along the trajectory. Hence, $P={(x,y,z):z>-1}$ is a positively invariant set. Then, as @H. H. Rugh has already said, we can use the fact that $L$ exponentially decreases to zero for $z>-1$, thus, any solution starting at $x_0\in P$ tends to zero. – AVK Dec 21 '17 at 21:04
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I will add a more detailed answer to the hint given by @H. H. Rugh (credit goes to him).

As H. H. Rugh proposed you need to find a Lyapunov function and the proposed function works. So we start by using

$$L(x,y,z)=x^2+2y^2+z^2$$

as a positive definite function, which is only zero for the origin (which is an equilibrium of the system).

The time derivative is given by

$$\dfrac{d}{dt}L(x,y,z)=2x\dot{x}+4y\dot{y}+2z\dot{z}$$ $$=2x(\varepsilon x+2y)(z+1)+4y(\varepsilon y-x)(z+1)+2z(-z^3)$$ $$=-2(-\varepsilon)(z+1)(x^2+2y^2)-2z^4.$$

As can be easily seen by the last equation we can see that the time derivative is negative definite as long as $\varepsilon<0$ and $z+1>0 \implies z>-1$.

In order to estimate the region of attraction, we observe that $z>-1$ and $L(x,y,z)=x^2+2y^2+z^2<c$ in which we need to find the largest $c \in \mathbb{R}$ such that it does not violate $z>-1$. The Set $x^2+2y^2+z^2<c$ is an ellipsoid which is squeezed in the $y$-direction by a factor of 2 (draw a picture of the situation to conclude the following statement). Hence, we can conclude that $c=1$ qualifies for the region of attraction.

MrYouMath
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  • Why $L(x,y,z)=x^2+2y^2+z^2<c?$ with this are you saying that $L$ is bounded? – user441848 Dec 21 '17 at 13:22
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    It works for all of $z>-1$. You don't need the restriction on $x,y$. Just note that $L$ is non-increasing here and that after a time $\delta$, you have $z>-1/2$, say. From then on $\dot{L}\leq -c L$ with $c>0$, so $L$ tends exponentially fast to zero. – H. H. Rugh Dec 21 '17 at 15:34
  • @H.H.Rugh what restriction on $x,y$? – user441848 Dec 21 '17 at 20:15
  • @Anneliset. None. Whatever starting values of $x,y$ and $z>-1/2$, $L$ goes monotonically to zero. Even more, for any $x,y,z$ the value of $L$ will eventually go to zero (through not necessarily monotonically), so every point in the space is in the basin of attraction. [The latter is a bit harder to show] – H. H. Rugh Dec 21 '17 at 20:38
  • @H.H.Rugh I don't see how $L$ goes to zero since $L(x,y,z)=x^2+2y^2+z^2>0$ is positive. With the additional condition of $z>-1/2$ – user441848 Dec 21 '17 at 20:47
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The equation for $z$ is independent of $x$ and $y$, so let's look at that first. Since $z^\prime$ is positive when $z$ is negative (and vice versa) and $z^\prime = 0$ when $z=0$, we know that

1) $z\rightarrow 0$ as $t\rightarrow \infty$

2) If $z_0 > -1$, then $z > -1$ for all $t>0$ (and so $z+1>0$)

3) $z(t)$ can be explicitly solved for

We let $A(t) = z(t)+1$, which is guaranteed to be a strictly positive function for all $t\geq 0$ as long as $z_0 > -1$. This reduces the first two equations to a non-autonomous linear system:

$\begin{bmatrix} x \\ y \end{bmatrix}^\prime = -A(t)\begin{bmatrix} \delta & -2 \\ \delta & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$

where $\delta = -\epsilon > 0$. From here you just need to show that any initial values $(x_0,y_0) \rightarrow (0,0)$ when $t\rightarrow \infty$.

Alex Jones
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