Does Asymptotic Stability Imply the Existence of a Lyapunov Function for a Nonlinear System

lyapunov-functionsnonlinear dynamicsstability-in-odes

For a linear time-invariant system $\dot x = Ax,$ the inverse Lyapunov theorem asserts that if the origin is asymptotically stable, then a Lyapunov function in the form $V(x) = x^\top P x$ for some positive definite function $P.$

Is there a similar result for nonlinear systems? Namely, for a nonlinear dynamical system in the form $\dot x = f(x)$ such that the origin $x=0$ is globally asymptotically stable, is the existence of a Lyapunov function guaranteed? (Of course, even if the Lyapunov function exists, it might be very tricky to find.) If the existence of a Lyapunov function is not guaranteed, then are there any known counter-examples?

Best Answer

Yes, it is Theorem 4.17 in Khalil's book:

Let $x=0$ be an asymptotically stable equilibrium of $\dot{x}=f(x)$. Then there is a smooth positive definite function $V(x)$ and a continuous positive definite function $W(x)$ such that: $$ \frac{\partial V}{\partial x}\,f(x)=-W(x) $$

(I have omitted some details)

Related Solutions

Nonlinear System Stability using Lyapunov

If you want to investigate the local stability of an equilibrium point of a nonlinear system it is usually easier to linearize. If all eigenvalues of linearized system have a negative real part then that equilibrium point is (asymptotically and exponentially) stable and if any of the eigenvalues has a positive real part it is unstable. Only in the edge case, when it is not unstable but there are eigenvalues with zero real part, would it be required to consider more of the nonlinear dynamics in order to identify whether or not the equilibrium point is locally stable.

It can be noted that if you are able to show that the system is stable near the equilibrium point $x_{eq}$ by using the linearization

$$ \dot{e} \approx A\,e $$

with $e = x - x_{eq}$ and $\|e\|$ close to zero, you can also always find a Lyapunov function by solving the (continuous) Lyapunov equation

$$ A^\top P + P\,A = -Q, $$

for any given positive definite $Q=Q^\top$ the solution for $P$ is also positive definite. Namely, the associated Lyapunov function is given by $V(e) = e^\top P\,e$, for which it holds that $\dot{V}(e) \approx -e^\top Q\,e$ when $\|e\|$ is close to zero. A key thing to note here is that in general it is not true that $\dot{V}(e) \nleq 0\ \forall\, e$. Namely, if it would hold for all $e$ then you would have shown global instead of local stability. One final note is that the set of all $e$ such that $\dot{V}(e) < 0$ is a lower bound of the basin of attraction of that equilibrium point.

Lyapunov Function for Nonlinear System

With the help of MATHEMATICA.

$$ A = \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -k & -k-K & 0 & 0 \\ k & -k & 0 & -\delta \\ \end{array} \right) $$

and the command

P = LyapunovSolve[A, Transpose[A], -IdentityMatrix[4]]

we obtain

$$ P = \left[ \begin{array}{cccc} -\frac{\delta ^2 \left(3 k K+k (3 k+2)+K^2+K\right)+K (k+K) (2 k+K+1)}{2 \delta k (k+K) (2 k+K)} & -\frac{\delta ^2 k+\delta ^2 K+K}{4 \delta k^2+2 \delta k K} & -\frac{1}{2} & -\frac{k+1}{2 (k+K)} \\ -\frac{\delta ^2 k+\delta ^2 K+K}{4 \delta k^2+2 \delta k K} & \frac{\delta ^2 (k+K)+K (2 k+K+1)}{2 \delta (k+K) (2 k+K)} & \frac{k+1}{2 (k+K)} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{k+1}{2 (k+K)} & -\frac{\frac{\delta ^2 \left(2 k^2+2 k K+k+K^2\right)}{k+K}+(k+1) K}{2 \delta k} & \frac{K}{2 \delta } \\ -\frac{k+1}{2 (k+K)} & -\frac{1}{2} & \frac{K}{2 \delta } & \frac{k K+K}{2 \delta k+2 \delta K} \\ \end{array} \right] $$

Best Answer

Related Solutions

Nonlinear System Stability using Lyapunov

Lyapunov Function for Nonlinear System

Related Question