# How to show the state equation that $x_1=x_2$ can only happen at the origin

Take a look at this system: \begin{align} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= -\frac{x^2_1}{x_2} - x_2 + x_1 \end{align} Take a Lyapunov function as $$V(x_1,x_2) = x^2_1 + x^2_2$$ Its time derivative is $$\dot{V}(x_1,x_2) = -2(x_1 - x_2)^2 \leq 0$$ The authors state the following:

Since $$\dot{V}(x) = 0$$ for all $$x_1 = x_2$$, we need to check whether the origin is the only point where $$\dot{V}(x) = 0$$. It can be seen from the state equation that $$x_1 = x_2$$ can only happen at the origin, therefore the origin globally asymptotically stable.

It is not clear to me how from the state equation that $$x_1 = x_2$$ can only happen at the origin. Any suggestions!

Hint.

Making

$$\cases{ x_1\dot x_1 = x_1 x_2\\ x_2\dot x_2 = -x_1^2-x_2^2+x_1x_2 }$$

after subtracting we have

$$\frac 12(x_2^2-x_1^2)' = \frac 12((x_2+x_1)(x_2-x_1))'=-(x_1^2+x_2^2) < 0,\ \ \forall (x_1,x_2) \ne (0,0)$$

• "after subtracting" could you please illustrate this? Also, how did you come up with this term $\frac{1}{2}(x^2_2-x^2_1)'$. Commented Aug 16, 2019 at 9:00
• @CroCo $a - b$ is a subtraction and $\frac 12(x_2^2-x_1^2)' = x_2\dot x_2-x_1\dot x_1$ Commented Aug 16, 2019 at 9:35

I think that what the author states is badly worded. Namely I am assuming that the author is referring to LaSalle's invariance principle, which roughly states that when $$\dot{V}(x)=0$$ it won't remain zero unless $$x=0$$ this shows asymptotic stability. This can be done in your case by checking the dynamics at $$x_1=x_2$$ which allows us to simplify $$\dot{x}_2$$ to

$$\dot{x}_2 = -x_1.$$

Combining this with $$\dot{x}_1 = x_2$$ allows us to show that the system won't remain on the manifold $$x_1=x_2$$ unless $$x_1=x_2=0$$. This won't mean that the system will never have that $$x_1=x_2\neq0$$, but it does show that the system won't remain at that manifold.

• What do you mean by the term manifold?! Commented Aug 17, 2019 at 20:53
• @CroCo A lower dimensional subspace, in this case $\{x_1,x_2\in\mathbb{R}\,|\,x_1=x_2\}$. Commented Aug 18, 2019 at 11:33
• Thanks......... Commented Aug 18, 2019 at 11:34