Let's first define the stability of a singularity of a vector field-
Let, $f\in C^1(E)$, $E\subset \Bbb{R}^n$ be open and $x_0\in E$ is a singularity of $f$ i.e. $f(x_0)=\mathbf{0}$. Let, $\phi_t:E\to E$ be the flow generated by $f$. Then $x_0$ is said to be stable if $\forall > \epsilon>0,\exists \delta>0$ such that $\forall x\in N_\delta(x_0)$ we have $|\phi_t(x)-x_0|<\epsilon\ \forall t\ge0$ and $x_0$ is said to be asymtotically stable if $\ \forall x\in N_\delta(x_0) > \lim_{t\to\infty}\phi_t(x)=x_0$. We say $x_0$ is unstable if it is not stable.
Here $(0,0)$ is the singularity of $f$ as $f(0,0)=(0,0)$
There is a result which says if $x_0$ is hyperbolic singularity having all eigenvalues with negative real part, then $x_0$ is asymptotically stable. And if $x_0$ is hyperbolic singularity having all eigenvalues with postive real part, then $x_0$ is unstable.
So we first check the eigen values of $Df(0,0)$.
I found that $Df(0,0)=\begin{pmatrix}
0&1\\
-1&0
\end{pmatrix}$. Eigen values of this matrix is $\pm i$, hence origin is the non-hyperbolic singularity having all eigen-values with 0 real part. And there is a theorem which says that if $x_0$ is stable then $Df(x_0)$ have eigenvalues with non-positive real part. So, $(0,0)$ may be a stable.
Here we are asked to find a Lyapunov function to investigate the stability of $(0,0)$.
A Lyapunov function of a singularity $x_0\in E$ for a vector field $f\in C^1(E)$ is a real valued fuction $V\in C^1(E)$ with the property that $V(x_0)=\mathbf{0}$ and $V(x)>0\ \forall x\in E\setminus \{x_0\}$.
Can anyone help me to solve the problem? Any kind of idea, help is highly appreciated.
Given
$$ V(x,y)=-x^3+x^2 \left(y-\frac{1}{2}\right)+x y^2-\frac{1}{2} y^2 (2 y+1) $$
we have
$$ V_x = -x - 3 x^2 + 2 x y + y^2\\ V_y = x^2 - y + 2 x y - 3 y^2 $$
so $V(x,y)$ is a movement constant and partitions all movement orbits.
Follows a plot showing the orbits around the origin and characterizing a center in a conservative system.