For $(x,y,z)\in S^2=\{\vec{x}\in\mathbb{R}^3: \lVert\vec{x}\rVert=1\}$, I am given the 3D-system $$ \begin{align*} x'&=x(-x+f(x,y,z))\\ y'&=y(x-y+f(x,y,z))\\ z'&=z(y-z+f(x,y,z)) \end{align*} $$ with $f(x,y,z)=x^3-xy^2+y^3-yz^2+z^3$.
One equilibrium (among others) is given by $$ E=(a,2a,3a),\quad a=\sqrt{1/14}. $$
I would like to show that $E$ is globally stable within the set $$ M:=\{(x,y,z)\in S^2: x,y,z>0\} $$ (In fact, $E$ is the only equilibrium in $M$.)
Is there a Lyapunov-function $V$?
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I tried the function
$$ V(x,y,z)=\alpha(x-a)^2+\beta(y-2a)^2+\gamma(z-3a)^2, $$ where $\alpha,\beta,\gamma>0$.
At least, one gets that $V$ is positive definite on $M$, i.e. $V(E)=0$ and $V(x,y,z)>0$ for all $(x,y,z)\in M\setminus\{E\}$.
The derivative is $$ \begin{align*} V'&=2\alpha(x-a)x'+2\beta(y-2a)y'+2\gamma(z-3a)z'\\ &=2\alpha(x-a)(-x^2+x f)+2\beta(y-2a)(xy-y^2+yf)+2\gamma(z-3a)(yz-z^2+zf). \end{align*} $$
Unfortunately, I do not see whether $V'$ is negative definite on $M$: It is clear that $V'(E)=0$ but not whether $V'<0$ on $M\setminus\{E\}$. Actually, I tend to think that the sign of $V’$ changes on $M$ and thus my $V$ is not the right choice.
Do you have an idea?
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