Step 2: Calculation of the right-hand side of the Wei-Norman equation

The derivation of the Wei-Norman equation is carried out in two steps. The product term in the right-hand side of (10) is first computed by invoking the LTP function wner, in which the basis elements $B_{I}$ need to be replaced by $g_I$, $I=0,1,2,01,02,102,0102$. Next, the coefficients corresponding to the basis elements $g_I$, on both sides of equation (10)-(3) are equated using the LTP function wnde. More precisely, the LTP function wner ought to be invoked with the following parameters: rhwne:=wner($r$,$k-1$,$B$, $B_g$, lbdt), where $r=7$ is the dimension of the basis $B_g$, $k=4$ is the degree of nilpotency, and lbdt is the list of linear dependencies . The resulting expression is:

$$\begin{eqnarray*} rhwne&{:=}& \dot{\gamma}_0 f_0 + \dot{\gamma}_1 f_1 + \dot{... ...mma}_5 \gamma_0 a + \dot{\gamma}_6) [[f_0, f_1], [f_0, f_2]] \end{eqnarray*}$$

The function wnde(rhwne,$r$,$B$,lbdt) is applied to the above result returning the matrix $\xi(\gamma)$ (see equation (12)) and the set of equations:

$$\begin{eqnarray*} v_0 &=& \dot{\gamma}_0\\ v_1 &=& \dot{\gamma}_1\\ v_2 ... ...\gamma}_4 \gamma_3 + \dot{\gamma}_5 \gamma_0 a + \dot{\gamma}_6 \end{eqnarray*}$$

The inversion of $\xi(\gamma)$ results in the following Wei-Norman equation:

$$\left [\begin{array}{c}\dot{\gamma}_0\\ \dot{\gamma}_1\\ \dot{\... ...ma}_3\\ \dot{\gamma}_4\\ \dot{\gamma}_5\\ \dot{\gamma}_6 \end{array}\right ] = \left [\begin{array}{ccccccc} 1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\\ 0... ...in{array}{c} v_0\\ v_1\\ v_2\\ v_3\\ v_4\\ v_5\\ v_6 \end{array}\right ]~$$

with $\gamma_i(0)=0$, $i=0,1,\ldots,6$. A feasible control $\bar{u}$ satisfying the inclusion (28) is found as follows. First, (29) is integrated symbolically over $[0,T]$ and solved with respect to the extended controls $v_i$, $i=0,\ldots,6$ evaluated at $T$ to yield a symbolic expression for the reachable set $\mathcal{R}(T,U^e(p))$, now given as a set of admissible coordinate values $\gamma(T,\bar{u})$ for the original system. Next, a control $\bar{u}$ is found by solving (28) using standard nonlinear programming techniques; see [26] for details of this calculation. Stabilization is achieved by repetitive solution of (28) as shown by simulation results in Figure 1 which correspond to an initial condition $x_0=[-0.1\ 0\ 0.2\ 0\ 0\ 0.1]^T$. These results were obtained using a quadratic Lyapunov function $V(x)=\frac{1}{2}\Vert x\Vert^2$ and piece-wise constant controls $\bar{u}$ consisting of at most five switching times in any interval of length $T=0.1$.

Figure 1: Results for the stabilization of the rigid body.