# 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

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