Polyharmonic spline

In control theory, a control-Lyapunov function $V(x,u)$ 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 $V(x)\to \mathrm{\infty }$ as $\left|x\right|\to \mathrm{\infty }$), and such that

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

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

$u^{*}(x)=\arg {\min }_u\mathrm{\nabla }V(x,u)\cdot f(x,u)$

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

$m(1+q^2)\ddot{q}+b\dot{q}+K_{0}q+K_1{q}^3=u$

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

$({\ddot{q}}_d-\frac{u-K_{0}q-K_1{q}^3-b\dot{q}}{m(1+q^2)}+\alpha \dot{e})=-\frac{\kappa }{2}(\dot{e}+\alpha e)$

Solving for $u$ yields the control law

$u=m(1+q^2)({\ddot{q}}_d+\alpha \dot{e}+\frac{\kappa }{2}r)+K_{0}q+K_1{q}^3+b\dot{q}$

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

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References

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See also

• Artstein's theorem
• Lyapunov optimization
• Drift plus penalty