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

Question

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!

Answer 1

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

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

Answer 2

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.