1
$\begingroup$

I have the Lyapunov function $V(x_1,x_2) = x_1^2 + x_2^2$ and the following system.

$$ \begin{aligned} \dot{x_1} &= -x_1 + x_2^2\\ \dot{x_2} &= -x_1 x_2 - x_1^2 \end{aligned} $$

In this case,

$$\dot{V}(x_1,x_2)=2x_1\cdot[-x_1+x_2^2]+2x_2\cdot[-x_1x_2-x_1^2]=-2x_1^2(1+x_2),$$

which according to my textbook means that $V$ is a weak Lyapunov function and that the system is asymptotically stable.

For another system

$$ \begin{aligned} \dot{x_1} &= -x_1^3\\ \dot{x_2} &= -x_1^2 x_2 \end{aligned} $$

we have

$$ \dot{V}(x_1,x_2)=2x_1\cdot(-x_1^3)+2x_2\cdot (-x_1^2x_2)=-2x_1^4-2x_1^2x_2^2=-2x_1^2(x_1^2+x_2^2) $$

and in my textbook, it says that this system isn't asymptotically stable. How can I see that $V$ is a weak Lyapunov function and that one system is asymptotically stable and the other isn't from this information?

$\endgroup$
1
  • 4
    $\begingroup$ "My textbook" is not very useful information. It would be much more relevant to say which book it is. Anyway, you should look up LaSalle's theorem. $\endgroup$ May 6, 2022 at 17:56

1 Answer 1

3
$\begingroup$

The system $$ \begin{cases} \displaystyle \frac{dx}{dt}=-x+y^2\\\\ \displaystyle \frac{dy}{dt}=-xy-x^2\\\\x(0)=x_0 \\\\ y(0)=y_0\end{cases} \tag{1} $$ has only $(x^*,y^*)=(0,0)$ and $(x^*,y^*)=(1,-1)$ as critical (or stationary or equilibrium) point.

Let us consider the function $$V(x,y)=x^2+y^2=\|(x,y)\|^2.$$

It follows that, if $(x(t),y(t))$ is a solution to $(1)$, then $$\displaystyle \frac{dV(x(t),y(t))}{dt}=\nabla V(x(t),y(t))^T \left(\displaystyle \frac{dx}{dt},\displaystyle \frac{dy}{dt}\right)=-2x(t)^2(1+y(t)).$$

This means that, if $y_0>-1$ in $(1)$, then $$\displaystyle \frac{dV(x(t),y(t))}{dt}<0,\quad x(t)\neq 0.$$ If $x(t_0)=0$ and $y(t_0)\neq 0$, to some $t_0\in \mathbb{R}$, then $$ \begin{cases} \displaystyle \frac{dx}{dt}\vert_{t=t_0}=y(t_0)^2>0\\\\ \displaystyle \frac{dy}{dt}\vert_{t=t_0}=0\end{cases}, $$ which implies that $x(t_0+h)>0$, to some $h\approx 0$.

It follows that $V(x(t),y(t))$ is a decreasing function, when $y_0>-1$, and $(x_0,y_0)\neq (0,0)$, that is, $V(x(t),y(t))\to 0$, as $t\to +\infty$, and the system is (locally) asymptotically stable around the origin $(0,0)$.

This also helps us to imagine how are the integral curvas of $(1)$, as follows:

enter image description here

Now let us consider the system $$ \begin{cases} \displaystyle \frac{dx}{dt}=-x^3\\\\ \displaystyle \frac{dy}{dt}=-x^2y\\\\x(0)=x_0\\\\y(0)=y_0\end{cases} \tag{2} $$

This system has all points of the form $(0,y)$ as critical points.

It follows that, if $(x(t),y(t))$ is a solution to $(2)$, then $$\displaystyle \frac{dV(x(t),y(t))}{dt}=-2x(t)^2(x(t)^2+y(t)^2)<0,\quad x(t)\neq 0.$$

This means that $V(x(t),y(t))$ is a decreasing function, when $x_0\neq 0$. But, if we choose $x_0=0$, then $(x(t),y(t))=(0,y(0))$, $V(x(t),y(t))=|y(0)|$, and the system $(2)$ is not asymptotically stable. This also helps us to imagine how are the integral curvas of $(2)$, as follows, except on line $(0,y)$:

enter image description here

You can find many discussions on this subject seraching for "\(V(u)>0\) Lyapunov function" On SearchOnMath, for instance.

$\endgroup$
2
  • 1
    $\begingroup$ You state that system $(1)$ only has $(x^*,y^*)=(0,0)$ as equilibrium, but it also has $(x^*,y^*)=(1,-1)$ as equilibrium point. $\endgroup$ May 9, 2022 at 15:00
  • $\begingroup$ Thanks. It needs just a little change in my answer. I will do it. $\endgroup$ May 9, 2022 at 16:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .