# Determine the stability of $(x,y)=(0,0)$

Determine the stability of $(x,y)=(0,0)$:

1/$$\bf{\begin{cases} & \mathrm{ } \dot{x}= -2x-y+2xy^2-3x^3\\ & \mathrm{ } \dot{y}= \dfrac{x}{3}-y-x^2y-7y^3 \end{cases} \tag {1}}$$

2/ $$\bf{\begin{cases} & \mathrm{ } \dot{x}= x-xy^4\\ & \mathrm{ } \dot{y}= y-x^2y^3 \end{cases} \tag {2}}$$

I'm trying to find a Liapunov function $V(x,y)=???$. But I have no solution! :( Anyone can post a few hints (like $V(x,y)=?$)

Any help will be appreciated.Thanks!

• You realize that you cannot always find one, correct? Do you know one exists for each of these cases? Regards Sep 23 '13 at 17:26

Hints:

If we look at phase portraits for the two systems, we have:  From these, we can see there is a single critical point for the first and it looks well behaved, so there is a chance we can find one.

Hint: Try $V(x, y) = ax^2 + by^2$ and see if you can resolve $a$ and $b$.

For the second one, there are a lot of things going on in the phase portrait due to five critical points. I doubt you will be able to find one for this as it is all over the map. Also, it is clear the CP is unstable by just looking at the direction fields, but you can do an eigenvalue analysis of the Jacobian at $(0,0)$ to also see that.

• Did you figure out a. with the hint? What did you get for $a$ and $b$? Sep 23 '13 at 18:22
• @kimtahe6: What book do these problems come from? Regards Sep 23 '13 at 21:24
• @ Amzoti: I don't know what book do these problems come from :( Because these are my homework (since my professor :) . He had read in my class and he didn't tell us it from the book. Sep 24 '13 at 4:50
• @ Amzoti: Why you said that "I may have given a bad hint on the other problem and am looking at if for V"? What does it mean? Sep 24 '13 at 13:38
• @kimtahe6: I was not totally awake yet! The hint was perfect. Recall, the claim I made is that you could get this to work for the first, since there is only ONE critical point and it is well behaved (even in that case, I am not sure it is always possible to find $V$). For the second one, there are five critical points and it will not be possible to find such a $V$ as the answer describes. Is this clear? Sep 24 '13 at 14:19

Solution:

For system of equations: $$\bf{\begin{cases} & \mathrm{ } \dot{x}= x-xy^4\\ & \mathrm{ } \dot{y}= y-x^2y^3 \end{cases} \tag {2}}$$

We use the Chetaev theorem to prove this. Let the function $V(x,y)$ have the form $$V(x,y)=x^2-y^2$$

This function is positive definite in the subdomain $\mathcal{V}$ in which the inequality $|x| > |y|$ holds.

We calculate the derivative $\dfrac{dV}{dt}$ by virtue of the system and define its sign in the subdomain $\mathcal{V}$.

$$\dot{V}(x,y)=2(x^2-y^2)=2V(x,y)$$

• It can be seen that the derivative $\dfrac{dV}{dt}$ is also positive definite in the subdomain $\mathcal{V}$ defined by the relation $|x| > |y|$.

• Besides that, the function $V(x,y)$ is zero on the boundary of $\mathcal{V}$ including the point $(0,0)$.

So, all conditions of the Chetaev theorem are fulfilled. Therefore, the zero solution of the system is unstable.

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Is that ok? Amzoti