Construct a Liapunov function for the system (Determine the stability of $x \equiv 0$):

I have an example:$$\begin{cases} & \mathrm { } \dot{x}= -x^3+xy^2\\ & \mathrm { } \dot{y}= -2x^2y-y^3 \end{cases} \tag{1}$$

Here's my solution:

  • Let's try $V(x,y)=ax^2+by^2$.

Then we have: $\dot{V}(x,y)=-2ax^4+2(a-2b)x^2y^2-2by^4$

  • When $a-2b<0$, for instance $a=b=1$. We have $V(x,y)=x^2+y^2$ such that: $$V(0,0)=0,V(x,y)>0, \forall (x,y) \ne (0,0)\ \text{and} \ \dot{V}(x,y)<0$$

  • Hence, $x=y=0$ is asymptotically stable.


What about the system:

$$\begin{cases} & \mathrm{ } \dot{x}= y-3x^3\\ & \mathrm{ } \dot{y}= -x-7y^3 \end{cases} \tag{2}$$

How can we construct a Liapunov function for this system (Determine the stability of $x \equiv 0$ of the system).

I'm sorry I fixed it!

  • $\begingroup$ If one system is $(1)$, the other one should be $(2)$ $\endgroup$ – Robert Israel Sep 22 '13 at 4:20
  • $\begingroup$ Yes, $(2)$ Hihi :) It's a typo! ** Robert Israel** $\endgroup$ – kimtahe6 Sep 22 '13 at 4:23

I doubt that there is a closed-form global Liapunov function. Note that besides the centre at $(0,0)$ you have critical points at $(\pm \sqrt{3}/3,\mp 7 \sqrt{3}/9)$, which I think are saddle points.

EDIT: It looks like e.g. $$ V(x,y) = {x}^{2}+{y}^{2}+ 8.75\,x{y}^{3}+3\,{x}^{2}{y}^{2}+ 5.25\,{x}^{3}y$$ is a Liapunov function for $(x,y)$ near $(0,0)$.

EDIT: How could I find this? By symmetry, I wanted something invariant under $ x \to -x,\; y \to -y$, so terms of even total degree in $x$,$y$. So start with $x^2 + y^2 + \sum_{i=0}^4 a_i x^i y^{4-i}$. The lowest-order terms of $\dot{V}$ are of total degree $4$ in $x,y$. I then substituted $x = \cos(s)$, $y = \sin(s)$ into those terms, and chose $a_0, \ldots, a_4$ so that the result was negative for all $s \in [0,2\pi]$. The one above is far from the only possible choice (and I don't remember exactly why I chose that particular one).

  • $\begingroup$ You mean we consider the system of equation: $$\begin{cases} & \mathrm{ } 0= y-3y^3\\ & \mathrm{ } 0= -x-7y^3 \end{cases}; $$ Whence, $$(x,y) \in \{(0,0);(-\dfrac{7}{3\sqrt{3}},\dfrac{1}{\sqrt{3}}), (\dfrac{7}{3 \sqrt{3}},-\dfrac{1}{\sqrt{3}}) \}$$. And what We're going to do? You can tell me more? Plz! $\endgroup$ – kimtahe6 Sep 22 '13 at 9:53
  • $\begingroup$ Wow! OMG It's too long! Hihi :) I'm sorry when I have a question: how do you find it? Plz! $\endgroup$ – kimtahe6 Sep 23 '13 at 16:45
  • $\begingroup$ It's great! You're very intelligent! Robert Israel. I'll check it, but I think that you're correct! :) $\endgroup$ – kimtahe6 Sep 24 '13 at 4:43
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    $\begingroup$ As I said, it's only a local minimum, reflecting the fact that $(0,0)$ is only a local attractor, not a global attractor. $\endgroup$ – Robert Israel Oct 6 '13 at 6:48
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    $\begingroup$ That makes it much less interesting. You just need a quadratic. $\endgroup$ – Robert Israel Oct 9 '13 at 14:42

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