I have the following system: $$\dot{x_1} = x_2(x_3-2) \\ \dot{x_2} = x_1(x_3-2) \\ \dot{x_3}=-x_3^3 $$ and I want to determine its equilibrium points together with their stability. To find the equilibrium points, I wrote $$x_2=0 \text{ or } x_3=2 \\ x_1=0 \text{ or } x_3=2 \\ x_3 =0$$
Thus, only equilibrium point $x^*$ for the system is the origin $0$. I found $\det{Df(0)} = 0$ so I expect that the origin is a non-hyperbolic equilibrium point, probably asymptotically stable. To determine its type, I want to find a Liapunov function. Now, I assumed I can take $V(x)=c_1x_1^2 +c_2x_2^2+c_3x_3^3$ for some positive constants $c_i$. We have $V(0)=0$ and $V(x)>0$ for any $x \neq 0$. Then, $$\dot{V}(x)=\nabla V\cdot f(x) = 2c_1x_1x_2(x_3-2)+2c_2x_1x_2(x_3-2)-2c_3x_3^4.$$
$(\star )$ Now, let $c_1=c_2=1$ and $c_3 =2$ so that we get $$\dot{V}(x)=4x_1x_2(x_3-2)-4x_3^4.$$
However, I do not know how to determine the sign of this expression to conclude the result. I want to ask
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Is it enough to find a Lyapunov function which is always positive or negative to say that $x^*$ is stable/unstable/asymptotically stable?
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Similar to the first question but am I allowed to set arbitary values for $c_i$'s? (The $\star$ part)
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How can I deduce the result?
Best Answer
If the system is stable, then there exists a ball centered at the origin such that all trajectories starting in this ball are bounded. Let us consider the initial condition $$x(0) = \begin{bmatrix}x_{1}(0) \\ x_{2}(0) \\ 0\end{bmatrix}.$$ From the system's dynamics, it follows that for these initial conditions $x_3(t)\equiv 0$ for all $t\ge0$. Then the dynamics of $x_1(t)$, $x_2(t)$ is $$\begin{bmatrix}\dot{x}_{1}(t) \\ \dot{x}_{2}(t) \end{bmatrix} = \begin{bmatrix}0 & -2 \\ -2 & 0\end{bmatrix}\begin{bmatrix}{x}_{1}(t) \\ {x}_{2}(t) \end{bmatrix}.$$ This linear system is unstable, and thus all trajectories with nonzero initial conditions such that $x_3(0)=0$ are unbounded. The system is unstable.