# proving asymptotic stability dynamical system

I want to show the origin of the dynamical system

\begin{align} \dot{x}_1 &= -2x_1+x_2+x_1^3x_2^2\\ \dot{x}_2 &= -x_1-2x_2+x_1^2x_2^3 \end{align}

is asymptotically stable over an invariant set $D\subset\mathbb{R}^2$. We can write this system as

$$\dot{x} = Ax + g(x) = \begin{pmatrix}-2&1\\-1&-2\end{pmatrix}x + x_1^2x_2^2x.$$

Where $x=(x_1,x_2)$. Using the quadratic form $V(x)=x^TPx$ with

$$P = \begin{pmatrix}1/4 & 0\\0& 1/4\end{pmatrix}\implies \lambda_{min}(P) = 1/4, \,\, \Vert P\Vert = \lambda_{max}(P)=1/4$$

Where $P$ is given by the formula $A^TP+PA=-I$. It can be shown that $\dot{V}$ may be written as

\begin{align} \dot{V}(x) &= -\vert x\vert^2 + 2x^TPg(x)\\ &\le -\vert x\vert^2 + 2\vert x\vert \Vert P\Vert \vert g(x)\vert \end{align}

Now note that

$$\frac{\vert g(x)\vert}{\vert x\vert} = x_1^2x_2^2 \le \frac{1}{2}(x_1^2+x_2^2)^2 = \frac{1}{2}\vert x\vert^4<\varepsilon \iff \vert x\vert < (2\varepsilon)^{1/4}$$

Thus

$$\dot{V}(x)< -\vert x\vert^2 + 2\varepsilon\vert x\vert^2 \Vert P\Vert = -\vert x\vert^2[1-2\varepsilon\Vert P\Vert]<0\iff \varepsilon<1/(2\Vert P\Vert)$$

Recall that $P$ is positive definite so

$$\lambda_{min}(P)\vert x\vert^2 \le x^TPx = V(x)$$

We also have that $\vert x\vert < (2\varepsilon)^{1/4}<1/\Vert P\Vert^{1/4}$ so

$$\lambda_{min}(P)\vert x\vert^2 < \lambda_{min}(P)/\Vert P\Vert^{1/2}$$

Since there is a case where $V(x)=\lambda_{min}(P)\vert x\vert^2$ we have that

$$\lambda_{min}(P)\vert x\vert^2 \le V(x) < \lambda_{min}(P)/\Vert P\Vert^{1/2}$$

Therefore, the dynamical system is asymptotically stable on the invariant set

$$D = \left\{x\in\mathbb{R}^2\,\colon V(x) < \frac{\lambda_{min}(P)}{\Vert P\Vert^{1/2}}\right\} = \left\{x\in\mathbb{R}^2\,\colon V(x) < \frac12\right\}$$

However my lecturer has

$$D = \left\{x\in\mathbb{R}^2\,\colon V(x) < 2\right\}$$

What have I done wrong?

• Sure your lecturer does not use $V(x)=|x|^2$ instead of your $V(x)=\frac14|x|^2$? (I would, if I were her...) – Did Apr 6 '16 at 5:29
• @Did nope she has used $V(x)=x^TPx$ with the same $P$ I have above. – user2850514 Apr 6 '16 at 10:12

## 1 Answer

No idea why this answer is downvoted. Let me urge readers misled by the OP's refusal that the conundrum where they are lost is simply due to a confusion between the choices $V(x)=|x|^2$ and $V(x)=\frac14|x|^2$, to reconsider, since this confusion is obviously the correct explanation.

This might be due to some confusion in the definition of $V$ since $x=\left(\frac32,\frac32\right)$ is not in the invariant set of this dynamical system although, with your choice of $P$, $x^TPx=\frac14|x|^2=\frac98<2$. Note that $P=\frac14I$ hence $x^TPx$ is just a fancy way of writing $\frac14x^Tx=\frac14|x|^2$. On the other hand, the (more natural) choice $V(x)=|x|^2$ yields your lecturer's answer.

In support of this explanation, note that $D=\{x\in\mathbb R^2\mid x^TPx<\frac12\}=\{x\in\mathbb R^2\mid |x|^2<2\}$ is indeed an invariant domain.

• Hi, the definition of $V$ asked to use is $V(x) = x^TPx$ where $P$ is defined above in the OP. – user2850514 Apr 9 '16 at 13:58
• @user2850514 We know that, you already explained it and it was discussed in the comments, maybe rehashing it once again is not so useful? – Did Apr 9 '16 at 18:47
• but you have not changed your answer. I thought maybe you forgot so I was reminding you. – user2850514 Apr 10 '16 at 16:08
• Yeah, and since I explained everything sensible there was to say on the subject, there is no point in "changing my answer". Do you understand why? Anyway, if you are waiting for another explanation, do not hold your breath... – Did Apr 10 '16 at 16:11
• Ok. Thank you for your response! My apologies for any inconvenience. – user2850514 Apr 10 '16 at 16:56