# What is a weak Lyapunov function and how can I use one to tell if a system is asymptotically stable?

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?

• "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. May 6, 2022 at 17:56

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:

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 $$(0,y)$$:

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

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