# Demonstrating Global Asymptotic Stability of a point

I'm trying to determine whether the origin of the following system is globally asymptotical stable (g.a.s.) or not: $$\begin{gathered} \dot{x}_1 = -x_1-x_2 \\ \dot{x}_2 = -x_1-x_2^3. \end{gathered}$$ In order to have g.a.s., the origin must be the unique equilibrium point of the system, and also it must be asymptotic stable.\ It is simple to notice that the system has a unique equilibrium point in the origin.\ I'm trying to demonstrate the asymptotic stability of the origin by selecting the following Lyapunov Function: $$V = x_1^2/2 + x_2^2/2$$ Taking its time derivative and substituting the original system: $$\dot{V} = -x_1^2 -x_2^4 -2x_1x_2$$ But it seems very difficult to demonstrate that this function is always strictly less than $$0$$. Any suggestions or hints would be highly appreciated

• Have you already checked the linearization around the origin? Commented Dec 25, 2022 at 22:38
• @KwinvanderVeen of course. The eigenvalues are negatives and equal to -1 Commented Dec 25, 2022 at 22:57

You have the system

\begin{align} \dot{x}_1&=-x_1-x_2\\ \dot{x}_2&=-x_1-x_2^3 \end{align}

The linearization matrix of this system is

$$\begin{pmatrix}-1&-1\\-1&-3x_2^2\end{pmatrix}$$

so at the origin we have

$$\begin{pmatrix}-1&-1\\-1&0\end{pmatrix}$$

which has eigenvalues $$-\frac{1}{2} \pm \frac{\sqrt{5}}{2}$$, or approximately, $$0.6180$$ and $$-1.6180$$. Since both eigenvalues are different from zero and have a different sign (one strictly positive, one strictly negative), the origin is unstable.

Btw., note that the system cannot be globally stable because you have multiple (three) equilibrium points: $$(0,0),(1,-1),(-1,1)$$ all make $$\dot{x}=0$$, so even if the linearization would be stable, the system could not be globally asymptotically stable.

We can also evaluate the stability of one of the equilibrium points, say $$(1, -1)$$. Doing a coordinate transformation yields

$$p = x_{1} - 1 \rightarrow x_{1} = p + 1$$ $$q = x_{2} + 1 \rightarrow x_{2} = q - 1$$

and the dynamics of the transformed system becomes

$$\dot{p} = \dot{x}_{1} = - p - q$$ $$\dot{q} = \dot{x}_{2} = - p - 1 - (q - 1)^{3} = - p - 3 q + 3 q^{2} - q^{3}$$

If linearization is performed, then we have

$$\dot{p} = - p - q$$ $$\dot{q} = - p - 3 q$$

Consider the Lyapunov function:

$$V = \left[\matrix{p & q}\right] \left[\matrix{3 & -1 \cr -1 & 1}\right] \left[\matrix{p \cr q}\right] > 0 \quad \forall \; p \neq 0, q \neq 0$$ $$V = 3 p^{2} - 2 p q + q^{2}$$

Then

$$\dot{V} = 6 p \dot{p} - 2 \left(\dot{p} q + p \dot{q}\right) + 2 q \dot{q}$$ $$\dot{V} = 6 p (- p - q) - 2 (- p - q) q - 2 p (- p - 3 q) + 2 q (- p - 3 q)$$ $$\dot{V} = - 6 p^{2} - 6 p q + 2 p q + 2 q^{2} + 2 p^{2} + 6 p q - 2 p q - 6 q^{2}$$ $$\dot{V} = - (6 - 2) p^{2} - (6 - 2) q^{2}$$ $$\dot{V} = - 4 p^{2} - 4 q^{2} < 0 \quad \forall \; p \neq 0, q \neq 0$$

Thus, the equilibrium point $$(1, -1)$$ is local asymptotically stable. The stability of the other point $$(-1, 1)$$ can also be evaluated almost similarly.