Control-Lyapunov function

In control theory, a control-Lyapunov function (CLF) It was later shown by Francis H. Clarke that every asymptotically controllable system can be stabilized by a (generally discontinuous) feedback. Artstein proved that the dynamical system (2) has a differentiable control-Lyapunov function if and only if there exists a regular stabilizing feedback u(x).

Constructing the Stabilizing Input

It is often difficult to find a control-Lyapunov function for a given system, but if one is found, then the feedback stabilization problem simplifies considerably. For the control affine system (2), Sontag's formula (or Sontag's universal formula) gives the feedback law $k:{\mathbb{ℝ}}^n\to {\mathbb{ℝ}}^m$ directly in terms of the derivatives of the CLF. In the special case of a single input system $(m=1)$, Sontag's formula is written as

$k(x)=\{\begin{array}{ll}-\frac{L_{f}V(x)+\sqrt{{[L_{f}V(x)]}^2+{[L_{g}V(x)]}^4}}{L_{g}V(x)}& \text{ if }{L}_{g}V(x)\ne 0\\ 0& \text{ if }{L}_{g}V(x)=0\end{array}$

where $L_{f}V(x):=⟨\mathrm{\nabla }V(x),f(x)⟩$ and $L_{g}V(x):=⟨\mathrm{\nabla }V(x),g(x)⟩$ are the Lie derivatives of $V$ along $f$ and $g$, respectively.

For the general nonlinear system (1), the input $u$ can be found by solving a static non-linear programming problem

$u^{*}(x)=\underset{u}{\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{i}\mathrm{n}}\mathrm{\nabla }V(x)\cdot f(x,u)$

for each state x.

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

$r\mapsto V(r):=\frac{1}{2}{r}^2$

which is positive for all $r\ne 0$.

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

$V=V(0)\exp (-\kappa t)$

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:

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

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

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

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

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

See also

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