Determining Stability of Equilibrium Point for x'=y-x3-xy2, y'=-x-x2y-y3

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SUMMARY

The equilibrium point for the system defined by the equations x' = y - x³ - xy² and y' = -x - x²y - y³ is determined to be at (0, 0). Using the Lyapunov function V(𝑥) = (x₁² + x₂²)/2, it is established that V' (𝑥) = - (x₁² + x₂²)² ≤ 0 for all 𝑥, confirming that the system is stable. This analysis provides a definitive conclusion regarding the stability of the equilibrium point.

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onie mti
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i have this system

x'=y-x3-xy2
y'=-x-x2y-y3

i worked it out and found the equilibrium point to be 0.

how do i determine whether it is stable, assymp stable or not stable
 
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onie mti said:
i have this system

x'=y-x3-xy2
y'=-x-x2y-y3

i worked it out and found the equilibrium point to be 0.

how do i determine whether it is stable, assymp stable or not stable

Let's define $x_{1}$ and $x_{2}$ the state variables, so that the system becomes...

$\displaystyle x_{1}^{\ '} = x_{2} - x_{1}^{3} - x_{1}\ x_{2}^{2}$

$\displaystyle x_{2}^{\ '} = - x_{1} - x_{1}^{2}\ x_{2} - x_{2}^{3}\ (1)$

You can choose as Lyapunov function $\displaystyle V(\overrightarrow x)= \frac{x_{1}^{2} + x_{2}^{2}}{2}$, so that is...

$\displaystyle V^{\ '} (\overrightarrow x) = x_{1}\ x_{2} - x_{1}^{4} - x_{1}^{2}\ x_{2}^{2} - x_{1}\ x_{2} - x_{1}^{2}\ x_{2}^{2} - x_{2}^{4} = - (x_{1}^{2} + x_{2}^{2})^{2} \le 0\ \forall \overrightarrow x\ (2)$

... so that the system is stable...

Kind regards

$\chi$ $\sigma$
 

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