I have the following system of equations:

$ \begin{cases} \frac{du}{dt} = v - v^3 \,, \\ \frac{dv}{dt} = -u - u^3 \,. \end{cases} $

I'm asked to find a Lyapunov function (Lyapunov's second method) to determine the stability around the origin. Using a linearization near the origin, I have found that the eigenvalues of the Jacobian are $\pm i$ and hence, the origin is a stable center point.

I figured this means I need to find a positive definite function (that is zero in the origin) and has negative semidefinite derivative (with respect to the system).

The questions in the book $\textit{Elementary Differential Equations and Boundary Value Problems}$ by $\textit{Boyce}$ and $\textit{DiPrima}$ are usually solved by trying the polynomials $V(u,v) = au^2 + bv^2$ or $V(u,v) = au^2 + buv + cv^2$. Sometimes a change to polar coordinates is made to determine a radius in which the derivative is negative. But I can't seem to ensure a derivative that is less or equal to zero in this case, for example:

Take $V(u,v) := au^2 + bv^2$, then

$ \begin{align*} \dot V &= 2auu' + 2bvv' \\ &= 2au(v-v^3) + 2bv(-u-u^3) & \mbox{let (for example) $a=b=1$}\\ &= -2uv^3 - 2vu^3 \end{align*} $

As these are cubic terms, they may very well be positive.


The Lyapunov function in this example is a bit more complicated than you expect. This system of equations looks very nice: equation for $u$ contains only $v$ and vice versa. Sometimes it's useful to switch back from first order system of ODEs to first order ODE — especially when the first order ODE has a closed form solution. And because it looks nice we can suspect that there is such solution. So instead of system we get equation: $$\frac{dv}{du} = \frac{-u-u^3}{v-v^3}$$ or (in symmetric form) $$ (u+u^3)\, du + (v-v^3) \, dv = 0 .$$ This is an exact equation and its general solution can be written via the function $$\Phi(u, v) = \frac{u^2}{2} + \frac{u^4}{4} + \frac{v^2}{2} - \frac{v^4}{4}.$$

When you have an exact equation (or separable, which is a particular case of an exact equation), you can use this $\Phi(u, v)$ as a Lyapunov function. But what happens if we calculate the derivative of $\Phi(u, v)$ w.r.t. to a system of ODEs? Let's check: $$\frac{d}{dt} \Bigl ( \Phi(u(t), v(t)) \Bigr ) = \Phi'_{u} (u(t), v(t)) \cdot \dot{u}(t) + \Phi'_{v}(u(t), v(t)) \cdot \dot{v}(t) = $$ $$ = (u(t)+u^3(t))\cdot(v(t) -v^3(t)) + (v(t) - v^3(t)) \cdot (-u(t)-u^3(t)) \equiv 0. $$

Technically, we satisfy the conditions of Lyapunov theorem: being exactly $0$ satisfies $\leqslant 0$ and it means that the equilibrium at the origin is Lyapunov stable. But there's more to it: actually, this system posesses the first integral. It means that all trajectories of this system don't leave the level sets $\Phi(u, v) = C$. If your planar system has first integral, then equilibria with $\pm i \omega$ eigenvalues are always true center equilibria.

  • $\begingroup$ Are we not also supposed to find a neighborhood U of the origin in which $\Phi(U) > 0$? I figure that is equivalent to what you are saying in your last remark, yet; how do I make this concrete? I don't see how that follows from the properties of the constructed $\Phi$ yet. Thank you $\endgroup$ – Dennis van den Berg Apr 15 '17 at 3:00
  • $\begingroup$ Technically, you are right -- we have to construct such neighbourhood to apply Lyapunov theorem. And we can: observe that you can rewrite this function as $\Phi(u, v) = \frac{u^2}{2} + \frac{u^4}{4} + \frac{v^2}{2} ( 1 -\frac{v^2}{2})$: it's obvious that when $v^2 < 2$ function $\Phi(u, v) > 0$. I wasn't very careful with this part of requirements of Lyapunov theorem because the property of first integral is more useful here and tells you much more about dynamics. $\endgroup$ – Evgeny Apr 15 '17 at 6:06

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