Is there any Lyapunov function to investigate the stability of the following dynamical system? $$ dx/dt= y + x(1-x^2-y^2) $$ $$ dy/dt=-x + y(1-x^2-y^2) $$ My view: I know this system has a limit cycle since the closed path $x^2+y^2=1$ is approached spirally from both the inside and outside by nonclosed paths as t tends to infinity. But I cannot find any Lyapunov function.Please help.
Asked
Active
Viewed 1,880 times
1
-
But the system is NOT stable, is it? – Did Apr 09 '17 at 18:32
1 Answers
2
The phase portrait a good place to start.
Candidate Lyapunov function $$ V(x,y) = \frac{1}{2} \left( x^{2} + y^{2} \right) $$ Is this a strong Lyapunov function? Criteria:
- $V(x,y)$ is positive definite about the point $(x,y) = (0,0)$
- $\dot{V}$ is negative semidefinite about the point $(x,y) = (0,0)$
Is $V(x,y)$ positive semidefinite about the origin? Compute the discriminant $$ \det \left[ \begin{array}{cc} V_{x,x} & V_{x,y} \\ V_{y,x} & V_{y,y} \end{array} \right] = \det \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = 1 $$ Because the discriminant is positive, the function $V(x,y)$ is positive definite.
The derivative $$ \dot{V}(x,y)= \nabla{V} \cdot \left[ \begin{array}{c} \dot{x} \\ \dot{y} \end{array} \right] = \left(x^{2}+y^{2}\right) \left(1 - x^{2} - y^{2} \right) $$ positive inside the unit circle, and negative outside.
So, no, $V(x,y)$ is not a strong Lyapunov function.
@Did: Red lines: nullclines where $x'=0$; yellow $y'=0$. Thanks for checking, $\dot{V}$ is corrected.
dantopa
- 10,768