Lyapunov stability

Question

i have a question regarding Lyapunov stability and basin of attraction.
Let

$${x}'=-x-y$$ $${y}'=2x-y+y^3$$ Use $$V(x,y)=x^{2}+\frac{1}{2}y^{2}$$ to determine the stability of (0,0) and a basin of attraction.
We have $$\dot{V}(x,y)=2x(-x-y)+y(2x-y+y^3)$$ or $$\dot{V}(x,y)=-2x^2-y^2+y^4$$. Now

1. V is positive definite and continuously differentiable
2. V(0,0)=0

Now,$$\dot{V}(x,y)$$ is negative definite if $$-y^2+y^4<0$$ or
$$-1 This is where i stuck ,i don't understand the last step from which he concluded that $$c\in (0,1/2)$$.Can someone help me?

Answer 1

$$ % f(x) = % \left[ \begin{array}{c} \dot{x} \\ \dot{y} \end{array} \right] % = % \left[ \begin{array}{c} -x -y \\ 2x - y + y^{3} \end{array} \right] % $$

Before looking at the Lyapunov function, let's take a peek at the phase portrait. The red dashed line represents $\dot{x} = 0$, purple $\dot{y}=0.$

The critical points are the origin and $(\pm \sqrt{3}, \mp \sqrt{3})$

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Lyapunov function $$ V(x,y) = x^{2} + \frac{1}{2} y^{2} $$ Gradient of Lyapunov function $$ \nabla V = \left[ \begin{array}{c} 2x \\ y \end{array} \right] $$ Time derivative $$ \dot{V} = \nabla V \cdot f = \left[ \begin{array}{c} 2x \\ y \end{array} \right] \cdot \left[ \begin{array}{c} -x -y \\ 2x - y + y^{3} \end{array} \right] = -2x^{2} - y^{2} + y^{4} $$ This function is shown below. The red contour line is the $\dot{V}=0$ contour. The region between these two red lines is the region where $\dot{V}<0$. The band for $-1 < y < 1$ is now apparent.