Exercise :
Consider the dynamical system : $$x' = -x +y^3$$ $$y' = -y+ax^3$$ with $(x,y) \in \mathbb R^2$. Find the stationary point of the system and study their stability for every value of $a$. Show that if $a=1$ and $V(x,y) \leq R$, where $R>0$ is a constant and $V(x,y) = \frac{1}{2}x^2 + \frac{1}{2}y^2$, then the derivative $V'(x,y)$ along the solutions of the dynamical system, satisfies the inequality $V'(x,y) \leq -2(1-R)V(x,y)$ and finally, estimate a stability area of $(0,0)$ for the case of $a=1$ using the Lyapunov Functional given.
Attempt / Question :
For the first part, I've elaborated a complete solution, as one can easily find the stationary points from solving the system :
$$\begin{cases} -x+y^3 \space\space=0\\ -y + ax^3=0\end{cases}$$
which yields $3$ different stationary points $(x,y)$.
For the stability and the kind of the stationary points, it's enough to find the eigenvalues for each stationary point for the linearisation matrix (jacobian), which is :
$$J(x,y) = \begin{bmatrix} -1 & 3y^2 \\ 3ax^2 & -1 \end{bmatrix}$$
For the part of the question in which I have an issue now, regarding the next segment of the problem :
The functional derivative, is given as :
$$\dot{V}(x,y) = \nabla V(x)f(x,y) = V_xf_x + V_yf_y = 2x(-x+y^3)+2y(-y+x^3) $$
$$\Leftrightarrow$$
$$\dot{V}(x,y) = -2x^2 + 2xy^3 - 2y^2 + 2yx^3$$
$$\Leftrightarrow$$
$$\dot{V}(x,y) = -2x^2 - 2y^2 + 2xy(y^2 + x^2)=-4V(x,y) + 4xyV(x,y)= 4V(x,y)(xy-1)$$
From this point on though, how would one proceed to show the inequality asked ? $$\dot{V}(x,y) \leq -2(1-R)V(x,y)$$
For the final part, I guess you just want to set a domain for the constant $R$ such that the functional is negative, but correct me if I'm wrong.
If the Lyapunov function is as in the citation ($V=\frac12 x^2+\frac12 y^2$, not $x^2+y^2$), then the derivative is $$ \dot V=x(-x+y^3)+y(-y+x^3)=-x^2-y^2+xy(x^2+y^2) $$ $$ =(x^2+y^2)(xy-1)=2(xy-1)V; $$ but $$\left(x-y\right)^2=x^2-2xy+y^2\ge0\;\Rightarrow \; xy\le \frac{x^2+y^2}2=V\le R;$$ finally, $$ \dot V\le 2(R-1)V=-2(1-R)V. $$
The estimation of the domain of attraction can be taken as the level set $$ \Omega=\left\{ (x,y)\in\mathbb R^2:\; V(x,y)= \frac12(x^2+y^2)<1 \right\} $$ (please notice the strict inequality, we need to exclude the equilibrium points $(-1,-1)$ and $(1,1)$ from the set). Indeed, for any $(x,y)\in \Omega$ we have $\dot V< 0$, except for the origin, thus, we can replicate the proof of the Lyapunov's asymptotic stability theorem to show that any solution starting in $\Omega$ tends to zero at $t\to+\infty$.