Control-Lyapunov function

In control theory, a control-Lyapunov function V(x,u) ^[1]is a generalization of the notion of Lyapunov function V(x) used in stability analysis. The ordinary Lyapunov function is used to test whether a dynamical system is stable (more restrictively, asymptotically stable). That is, whether the system starting in a state $x \ne 0$ in some domain D will remain in D, or for asymptotic stability will eventually return to x = 0. The control-Lyapunov function is used to test whether a system is feedback stabilizable, that is whether for any state x there exists a control u(x,t) such that the system can be brought to the zero state by applying the control u.

More formally, suppose we are given a dynamical system

$\dot{x}(t)=f(x(t))+g(x(t))\, u(t),$

where the state x(t) and the control u(t) are vectors.

Definition. A control-Lyapunov function is a function V(x,u) that is continuous, positive-definite (that is V(x,u) is positive except at x = 0 where it is zero), proper (that is $V(x)\to \infty$ as $|x|\to \infty$), and such that

$\forall x \ne 0, \exists u \qquad \dot{V}(x,u) < 0.$

The last condition is the key condition; in words it says that for each state x we can find a control u that will reduce the "energy" V. Intuitively, if in each state we can always find a way to reduce the energy, we should eventually be able to bring the energy to zero, that is to bring the system to a stop. This is made rigorous by the following result:

Artstein's theorem. The dynamical system has a differentiable control-Lyapunov function if and only if there exists a regular stabilizing feedback u(x).

It may not be easy to find a control-Lyapunov function for a given system, but if we can find one thanks to some ingenuity and luck, then the feedback stabilization problem simplifies considerably, in fact it reduces to solving a static non-linear programming problem

$u^*(x) = \arg\min_u \nabla V(x,u) \cdot f(x,u)$

for each state x.

The theory and application of control-Lyapunov functions were developed by Z. Artstein and E. D. Sontag in the 1980s and 1990s.

Example

Here is a characteristic example of applying a Lyapunov candidate function to a control problem.

Consider the non-linear system, which is a mass-spring-damper system with spring hardening and position dependent mass described by

$m(1+q^2)\ddot{q}+b\dot{q}+K_0q+K_1q^3=u$

Now given the desired state, q_d, and actual state, q, with error, e = q_d − q, define a function r as

$r=\dot{e}+\alpha e$

A Control-Lyapunov candidate is then

$V=\frac{1}{2}r^2$

which is positive definite for all $q \ne 0$, $\dot{q} \ne 0$.

Now taking the time derivative of V

$\dot{V}=r\dot{r}$

$\dot{V}=(\dot{e}+\alpha e)(\ddot{e}+\alpha \dot{e})$

The goal is to get the time derivative to be

$\dot{V}=-\kappa V$

which is globally exponentially stable if V is globally positive definite (which it is).

Hence we want the rightmost bracket of $\dot{V}$,

$(\ddot{e}+\alpha \dot{e})=(\ddot{q}_d-\ddot{q}+\alpha \dot{e})$

to fulfill the requirement

$(\ddot{q}_d-\ddot{q}+\alpha \dot{e}) = -\frac{\kappa}{2}(\dot{e}+\alpha e)$

which upon substitution of the dynamics, $\ddot{q}$, gives

$(\ddot{q}_d-\frac{u-K_0q-K_1q^3-b\dot{q}}{m(1+q^2)}+\alpha \dot{e}) = -\frac{\kappa}{2}(\dot{e}+\alpha e)$

Solving for u yields the control law

$u= m(1+q^2)(\ddot{q}_d + \alpha \dot{e}+\frac{\kappa}{2}r )+K_0q+K_1q^3+b\dot{q}$

with κ and α, both greater than zero, as tunable parameters

This control law will guarantee global exponential stability since upon substitution into the time derivative yields, as expected

$\dot{V}=-\kappa V$

which is a linear first order differential equation which has solution

V = V(0)e ^{− κt}

And hence the error and error rate, remembering that $V=\frac{1}{2}(\dot{e}+\alpha e)^2$, exponentially decay to zero.

If you wish to tune a particular response from this, it is necessary to substitute back into the solution we derived for V and solve for e. This is left as an exercise for the reader but the first few steps at the solution are:

$r\dot{r}=-\frac{\kappa}{2}r^2$

$\dot{r}=-\frac{\kappa}{2}r$

$r=r(0)e^{-\frac{\kappa}{2} t}$

$\dot{e}+\alpha e= (\dot{e}(0)+\alpha e(0))e^{-\frac{\kappa}{2} t}$

which can then be solved using any linear differential equation methods.

Notes

1. ^ Freeman (46)

References

• Freeman, Randy A.; Petar V. Kokotović (2008) (in English). Robust Nonlinear Control Design (illustrated, reprint ed.). Birkhäuser. pp. 257. ISBN 0817647589. http://books.google.com/books?id=_eTb4Yl0SOEC. Retrieved 2009-03-04. 

See also

• Artstein's theorem