Show that the equilibrium point is globally stable.

Question

Full question:

Consider the Lorenz model with $\dot{x} = \frac{3}{2}y-x$, $\dot{y} = \frac{1}{2}x-y-xz$, and $\dot{z} = xy-z$.

(a) Show that the Lorenz model has a unique equilibrium point. (b) Show that the equilibrium point is globally stable. (HINT: Use a Liapunov function.)

To clarify, I know part (a) and have found our equilibrium point. This can be done through some fairly simple algebra. However I'm not so skilled at using Liapunov functions yet, and would like to see how to handle part (b) while using one. My first attempt was using the function $V(x,y,z) = x^2+y^2+z^2$, but it didn't go so well.

Thanks in advance for any and all advice!

Answer 1

$V(x,y,z) = x^2 + y^2 + z^2$ will almost work. You should get $$ \dfrac{d}{dt} V(x,y,z) = 2 x \dot{x} + 2 y \dot{y} + 2 z \dot{z} = -(x-y)^2 - z^2 \le 0$$ Try instead $V(x,y,z) = x^2 + 2 y^2 + 2 z^2$.

Answer 2

Define $V(x,y,z)=ax^2+by^2+cz^2$. Let $U$ be a neighborhood of the origin.

Clearly $V(x,y,z)$ is continuously differentiable on the set $U-\{(0,0,0)\}$,

$V(0,0,0)=0$ and $V(x,y,z)>0 \,\, \forall(x,y,z)\in U-\{(0,0,0)\}$.

In order to show that $V(x,y,z)$ is a Lyapunov function and that the origin is asymptotically stable, you just need to find $a,b,c$ such that $\forall(x,y,z)\in U-\{(0,0,0)\}$

$$\nabla V(x,y,z)\cdot f(x,y,z)<0$$

Where $f(x,y,z)$ is the vector field.

Now you just need to find $a,b,c$ that satisfy this condition.