0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.