Control-Lyapunov function

In control theory, a control-Lyapunov function^[1] is a Lyapunov function $V(x)$ for a system with control inputs. 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 an autonomous dynamical system

\dot{x}=f(x,u)

where $x\in\mathbf{R}^n$ is the state vector and $u\in\mathbf{R}^m$ is the control vector, and we want to feedback stabilize it to $x=0$ in some domain $D\subset\mathbf{R}^n$ .

Definition. A control-Lyapunov function is a function $V:D\rightarrow\mathbf{R}$ that is continuously differentiable, positive-definite (that is $V(x)$ is positive except at $x=0$ where it is zero), and such that

\forall x \ne 0, \exists u \qquad \dot{V}(x,u)=\nabla V(x) \cdot f(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) \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 $\kappa$ and $\alpha$ , 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^{-\kappa t}

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

↑ Freeman (46)

References

Freeman, Randy A.; Petar V. Kokotović (2008). Robust Nonlinear Control Design (illustrated, reprint ed.). Birkhäuser. p. 257. ISBN 0-8176-4758-9. Retrieved 2009-03-04.

Control-Lyapunov function

Example

Notes

References

See also