Consider the following system of differential equations:
$$\begin{cases}
\frac{du}{dt}=-v\\
\frac{dv}{dt}=u-u^5
\end{cases}$$
I'm asked to determine the stability type of the isolated stationary point $(0,0)$. To do this, I first determined a linearization. The Jacobian is:
$$\mathbf{J}(u, v)=\begin{pmatrix}
0 & -1 \\
1-5u^4 &0
\end{pmatrix}$$
At $(0,0)$, this gives eigenvalues $r_1 = i$, $r_2=-i$. Given these eigenvalues, the type of stability can't be concluded.
I've then tried to apply a Lyapunov function. However, I can't seem to find an appropriate one:
- $V(u, v) = u^2 + v^2$ gives $\dot V(u, v) = -2u^5v$. While $V$ is positive definite, its derivative isn't definite in any way.
- $V(u, v) = au^2 + buv + cv^2$ gives $\dot V(u,v) = \big[bu^2+(2c-2a)uv-bv^2\big]-bu^6-2cu^5v$. This is also a dead end as far as I can see. I can't find any $a,b,c$ such that the part in square brackets is definite (and $V$ is as well).
I've tried to find an appropriate Lyapunov function online, but I can't seem to find one that would fit my case. Does anyone have a suggestion?
Best Answer
Suppose that the Lyapunov function is of the form $$ V(u,v)=au^2+bv^2+cu^m, $$ where $a,b,c,m$ are some numbers. The derivative is $$ \dot V= 2au(-v)+2bv(u-u^5)+mcu^{m-1}(-v) $$ $$ =2(b-a)uv-v(2bu^5+mcu^{m-1}). $$ It is equal to zero if $b=a$, $m=6$ and $mc=-2b$. On can take $a=b=3$, $c=-1$ and obtain $$ V(u,v)=3u^2+3v^2-u^6. $$ This function if positive if $|u|<\sqrt[4]3$ (except for the origin), therefore the origin is stable (but not asymptotially stable).